Measure theory
Table of Contents
Notation
 and are used to denote the indicator or characteristic function
Definition
Motivation
The motivation behind defining such a thing is related to the BanachTarski paradox, which says that it is possible to decompose the 3dimensional solid unit ball into finitely many pieces and, using only rotations and translations, reassemble the pieces into two solid balls each with the same volume as the original. The pieces in the decomposition, constructed using the axiom of choice, are nonmeasurable sets.
Informally, the axiom of choice, says that given a collecions of bins, each containing at least one object, it's possible to make a selection of exactly one object from each bin.
Measure space
If is a set with the sigmaalgebra and the measure , then we have a measure space .
Product measure
Given two measurable spaces and measures on them, one can obtain a product measurable space and a product measure on that space.
A product measure is defined to be a measure on the measurable space , where we've let be the algebra on the Cartesian product . This sigmaalgebra is called the tensorproduct sigmaalgebra on the product space, which is defined
A product measure is defined to be a measure on the measurable space satisfying the property
and
Let be a sequence of extended real numbers.
The limit inferior is defined
The limit supremum is defined
Premeasure
Given a space , and a collection of sets is an algebra of sets on if
 If , then
 If and are in , then
Thus, a algebra of sets allow only finite unions, unlike σalgebras where we allow countable unions.
Given a space and an algebra , a premeasure is a function such that
For every finite or countable collection of disjoint sets with , if then
Observe that the last property says that IF this "possibly large" union is in the algebra, THEN that sum exists.
A premeasure space is a triple where is a space, is an algebra, and a premeasure .
Complete measure
A complete measure (or, more precisely, a complete measure space ) is a measure space in which every subset of every null set is measurable (having measure zero).
More formally, is complete if and only if
If is a premeasure space, then there is a complete measure space such that
 we have
If is σfinite, then is the only measure on that is equal to on .
Atomic measure
Let be a measure space.
Then a set is called an atom if
and
A measure which has no atoms is called nonatomic or diffuse
In other words, a measure is nonatomic if for any measurable set with , there exists a measurable subset s.t.
πsystem
Let be any set. A family of subsets of is called a πsystem if
If , then
So this is an even weaker notion than being an (Boolean) algebra. We introduce it because it's sufficient to prove uniqueness of measures:
Theorems
Jensen's inequality
Let
 be a probability space
 be random variable
be a convex function Then is the supremum of a sequence of affine functions
for , with .
Then is welldefined, and
Taking the supremum over in this inequality, we obtain
be a convex function Then is the supremum of a sequence of affine functions
Suppose is convex, then for each point there exists an affine function s.t.
 the line corresponding to passes through
 the graph of lies entirely above
Let be the set of all such functions. We have
 because passes through the point
 beacuse all lies below
Hence
(note this is for each , i.e. pointwise).
Sobolev space
Notation
 is an open subset of
 denotes a infinitively differentiable function with compact support
is a multiindex of order , i.e.
Definition
Vector space of functions equipped with a norm that is a combination of norms of the function itself and its derivatoves to a given order.
Intuitively, a Sobolev space is a space of functions with sufficiently many derivatives for some application domain, e.g. PDEs, and equipped with a norm that measures both size and regularity of a function.
The Sobolev space spaces combine the concepts of weak differentiability and Lebesgue norms (i.e. spaces).
For a proper definition for different cases of dimension of the space , have a look at Wikipedia.
Motivation
Integration by parst yields that for every where , and for all infinitively differentiable functions with compact support :
Observe that LHS only makes sense if we assume to be locally integrable. If there exists a locally integrable function , such that
we call the weak th partial derivative of . If this exists, then it is uniquely defined almost everywhere, and thus it is uniquely determined as an element of a Lebesgue space (i.e. function space).
On the other hand, if , then the classical and the weak derivative coincide!
Thus, if , we may denote it by .
Example
is not continuous at zero, and not differentiable at −1, 0, or 1. Yet the function
satisfies the definition of being the weak derivative of , which then qualifies as being in the Sobolev space (for any allowed ).
Lebesgue measure
Notation
 denotes the collection of all measurable sets
Stuff
Given a subset , with the length of a closed interval given by , the Lebesgue outer measure is defined as
Lebesgue outermeasure has the following properties:
 Idea: Cover by .
(Monotinicy) if , then
Idea: a cover of is a cover of .
(Countable subadditivity) For every set and every sequence of sets if then
Idea: construct a cover of each , such that :
 Every point in is in one of the
Q: Is it possible for every to find a cover such that ? A: No. Consider . Given , consider . This is a cover of so . If is a cover by open intervals of , then there is at least one such that is a nonempty open interval, so it has a strictly positive lenght, and
If , then
Idea: , so . For reverse, cover by intervals giving a sum within . Then cover and by intervals of length . Put the 2 new sets at at the start of the sequence, to get a cover of , and sum of the lengths is at most . Hence,
If is an open interval, then .
Idea: lower bound from . Only bounded nonempty intervals are interesting. Take the closure to get a compact set. Given a countable cover by open intervals, reduce to a finite subcover. Then arrange a finite collection of intervals in something like increasing order, possibly dropping unnecessary sets. Call these new intervals and let be the number of such intervals, and such that
i.e. leftmost interval cover the startingpoint, and rightmost interval cover the endpoint. Then
Taking the infimum,
The Lebesgue measure is then defined on the Lebesgue sigmaalgebra, which is the collection of all the sets which satisfy the condition that, for every
For any set in the Lebesgue sigmaalgrebra, its Lebesgue measure is given by its Lebesgue outer measure .
IMPORTANT!!! This is not necessarily related to the Lebesgue integral! It CAN be be, but the integral is more general than JUST over some Lebesgue measure.
Intuition
 First part of definition states that the subset is reduced to its outer measure by coverage by sets of closed intervals
 Each set of intervals covers in the sense that when the intervals are combined together by union, they contain
 Total length of any covering interval set can easily overestimate the measure of , because is a subset of the union of the intervals, and so the intervals include points which are not in
Lebesgue outer measure emerges as the greatest lower bound (infimum) of the lengths from among all possible such sets. Intuitively, it is the total length of those interval sets which fit most tightly and do not overlap.
In my own words: Lebesgue outer measure is smallest sum of the lengths of subintervals s.t. the union of these subintervals completely "covers" (i.e. are equivalent to) .
If you take an a real interval , then the Lebesge outer measure is simply .
Properties
Notation
For and , we let
Stuff
The collection of Lebesgue measurable sets is a sigmaalgebra.
Easy to see is in this collection:
Closed under complements is clear: let be Lebesgue measurable, then
hence this is also true for , and so is Lebesgue measurable.
 Closed under countable unions:
Finite case: . Consider both Lebesgue measurable and some set . Since is L. measurable:
Since is L. measurable:
which allows us to rewrite the above equation for :
Observe that
By subadditivity:
Hence,
Then this follows for all finite cases by induction.
Countable disjoint case: Let , and . Further, let .
Hence is L. measurable. Thus,
Since the are disjoint and :
Let and note that . Thus, by indiction
Thus,
Taking :
Thus, is L. measurable if the are disjoint and L. measurable!
 Countable (notnecessarilydisjoint) case: If are not disjoint, let and let , which gives a sequence of disjoint sets, hence the above proof applies.
Every open interval is Lebesgue measurable, and the Borel sigmaalgebra is a subset of the sigmaalgebra of Lebesgue measurable sets.
Want to prove measurability of intervals of the form .
Idea:
 split any set into the left and right part
 split any cover in the same way
 extend covers by to make them open
is a measure space, and for al intervals , the measure is the length.
Cantor set
Define
For , with being identity, and
Let and . Then the Cantor set is defined
The Cantor set has a Lebesgue measure zero.
We make the following observations:
 Scaled and shifted closed sets are closed
 is a finite union of closed intervals and so is in the Borel sigmaalgebra
 σalgebras are closed under countable intersections, hence Cantor set is in the Borel σalgebra
 Finally, Borel σalgebra is a subset of Lebesgue measurable sets, hence the Cantor set is Lebesuge measurable!
Since Lebesgue measure satisfy for any Lebesgue measurable set with finite measure and any with . Since Lebesgue measure is subadditive, we have for any
Since , by induction, it follows that
Taking the infimum of over , we have that the Cantor set has measure zero:
Cardinality of the Cantor set
Let .
The terniary expansion is a sequence with such that
The Cantor set is uncountable.
We observe that if the first elements of the expansion for are in , then . But importantly, observe that some numbers have more than one terniary expansion, i.e.
in the terniary expansion. One can show that a number if and only if has a terniary expansion with no 1 digits. Hence, the Cantor set is uncountable!
One can see that if and only if terniary expansion with no 1 digits, since such an would land in the "gaps" created by the construction of the Cantor set.
Uncountable Lebesgue measurable set
There exists uncountable Lebesgue measurable sets.
Menger sponge
 Generalization of Cantor set to
Vitali sets
Let if and only if .
 There are uncountable many equivalence classes, with each equivalence class being countable (as a set).
 By axiom of choice, we can pick one element from each equivalence class.
 Can assume each representative picked is in , and this set we denote
Suppose, for the sake of contradiction, that is measurable.
Observe if , then there is a and s.t. , i.e.
Then, by countable additivity
where we've used
Hence, we have our contradiction and so this set, the Vitali set, is not measurable!
There exists a subset of that is not measurable wrt. Lebesgue measure.
Lebesgue Integral
The Lebesgue integral of a function over a measure space is written
which means we're taking the integral wrt. the measure .
Special case: nonnegative realvalued function
Suppose that is a nonnegative realvalued function.
Using the "partitioning of range of " philosophy, the integral of should be the sum over of the elementary area contained in the thin horizontal strip between and , which is just
Letting
The Lebesgue integral of is then defined by
where the integral on the right is an ordinary improper Riemann integral. For the set of measurable functions, this defines the Lebesgue integral.
Radon measure
 Hard to find a good notion of measure on a topological space that is compatible with the topology in some sense
 One way is to define a measure on the Borel set of the topological space
Let be a measure on the sigmaalgebra of Borel sets of a Hausdorff topological space .
is called inner regular or tight if, for any Borel set , is the supremum of over all compact subsets of of , i.e.
where denotes the compact interior, i.e. union of all compact subsets .
is called outer regular if, for any Borel set , is the infimum of over all open sets containing , i.e.
where denotes the closure of .
 is called locally finite if every point of has a neighborhood for which is finite (if is locally finite, then it follows that is finite on compact sets)
The measure is called a Radon measure if it is inner regular and locally finite.
Suppose and are two measures on a measures on a measurable space and is absolutely continuous wrt. .
Then there exists a nonnegative, measurable function on such that
The function is called the density or RadonNikodym derivative of wrt. .
If a RadonNikodym derivative of wrt. exists, then denotes the equivalence class of measurable functions that are RadonNikodym derivatives of wrt. .
is often used to denote , i.e. is just in the equivalence class of measurable functions such that this is the case.
This comes from the fact that we have
Suppose and are RadonNikodym derivatives of wrt. iff .
The δ measure cannot have a RadonNikodym derivative since integrating gives us zero for all measurable functions.
Continuity of measure
Suppose and are two sigmafinite measures on a measure space .
Then we say that is absolutely continuous wrt. if
We say that and are equivalent if each measure is absolutely continuous wrt. to the other.
Density
Suppose and are two sigmafinite measures on a measure space and that is absolutely continuous wrt. . Then there exists a nonnegative, measurable function on such that
Measurepreserving transformation
is a measurepreserving transformation is a transformation on the measurespace if
Measure
A measure on a set is a systematic way of defining a number to each subset of that set, intuitively interpreted as size.
In this sense, a measure is a generalization of the concepts of length, area, volume, etc.
Formally, let be a of subsets of .
Suppose is a function. Then is a measure if
Whenever are pairwise disjoint subsets of in , then
 Called σadditivity or subadditivity
Properties
Let be a measure space, and such that .
Then .
Let
Then , and by finite additivity property of a measure:
since by definition of a measure.
If are subsets of , then
We know for a sequence of disjoint sets we have
So we just let
Then,
Thus,
Concluding our proof!
Let be an increasing sequence of measurable sets.
Then
Let be sets from some .
If , then
Examples of measures
Let
 be a space
The δmeasure (at ) is
Sigmaalgebra
Definition
Let be some set, and let be its power set. Then the subset is a called a σalgebra on if it satisfies the following three properties:
 is closed under complement: if
 is closed under countable unions: if
These properties also imply the following:
 is closed under countable intersections: if
Generated σalgebras
Given a space and a collection of subsets , the σalgebra generated by , denoted , is defined to be the intersection of all σalgebras on that contain , i.e.
where
Let be a measurable space and a function from some space to .
The σalgebra generated by is
Observe that though this is similar to σalgebra generated by MEASURABLE function, the definition differs in a sense that the preimage does not have to be measurable. In particular, the σalgebra generated by a measurable function can be defined as above, where is measurable by definition of being a measurable function, hence corresponding exactly to the other definition.
Let and be measure spaces and a measurable function.
The σalgebra generated by is
Let be a space.
If is a collection of σalgebras, then is also a σalgebra.
σfinite
A measure or premeasure space is finite if .
A measure on a measure space is said to be sigmafinite if can be written as a countable union of measurable sets of finite measure.
Example: counting measure on uncountable set is not σfinite
Let be a space.
The counting measure is defined to be such that
On any uncountable set, the counting measure is not σfinite, since if a set has finite counting measure it has countably many elements, and a countable union of finite sets is countable.
Properties
Let be a of subsets of a set . Then
If , then
 If then
Borel sigmaalgebra
Any set in a topological space that can be formed from the open sets through the operations of:
 countable union
 countable intersection
 complement
is called a Borel set.
Thus, for some topological space , the collection of all Borel sets on forms a σalgebra, called the Borel algebra or Borel σalgebra .
More compactly, the Borel σalgebra on is
where is the σalgebra generated by the standard topology on .
Borel sets are important in measure theory, since any measure defined on the open sets of a space, or on the closed sets of a space, must also be defined on all Borel sets of that space.
Any measure defined on the Borel sets is called a Borel measure.
Lebesgue sigmaalgebra
Basically the same as the Borel sigmaalgebra but the Lebesgue sigmaalgebra forms a complete measure.
Note to self
Suppose we have a Lebesgue mesaure on the real line, with measure space .
Suppose that is nonmeasurable subset of the real line, such as the Vitali set. Then the measure of is not defined, but
and this larger set ( ) does have measure zero, i.e. it's not complete !
Motivation
Suppose we have constructed Lebesgue measure on the real line: denote this measure space by . We now wish to construct some twodimensional Lebesgue measure on the plane as a product measure.
Naïvely, we could take the sigmaalgebra on to be , the smallest sigmaalgebra containing all measureable "rectangles" for .
While this approach does define a measure space, it has a flaw: since every singleton set has onedimensional Lebesgue measure zero,
for any subset of .
What follows is the important part!
However, suppose that is nonmeasureable subset of the real line, such as the Vitali set. Then the measure of is not defined (since we just supposed that is nonmeasurable), but
and this larger set ( ) does have measure zero, i.e. it's not complete !
Construction
Given a (possible incomplete) measure space , there is an extension of this measure space that is complete .
The smallest such extension (i.e. the smallest sigmaalgebra ) is called the completion of the measure space.
It can be constructed as follows:
 Let be the set of all measure zero subsets of (intuitively, those elements of that are not already in are the ones preventing completeness from holding true)
 Let be the sigmaalgebra generated by and (i.e. the smallest sigmaalgreba that contains every element of and of )
 has an extension to (which is unique if is sigmafinite), called the outer measure of , given by the infimum
Then is a complete measure space, and is the completion of .
What we're saying here is:
 For the "multidimensional" case we need to take into account the zeroelements in the resulting sigmaalgebra due the product between the 1D zeroelement and some element NOT in our original sigmaalgebra
 The above point means that we do NOT necessarily get completeness, despite the sigmaalgebras defined on the sets individually prior to taking the Cartesian product being complete
 To "fix" this, we construct a outer measure on the sigmaalgebra where we have included all those zeroelements which are "missed" by the naïve approach,
Measurable functions
Let and be measurable spaces.
A function is a measurable function if
where denotes the preimage of the for the measurable set .
Let .
We define the indicator function of to be the function given by
Let . Then is measurable if and only if .
Let be a measure space or a probability space.
Let be a sequence of measurable functions.
 For each , the function is measurable
 The function is measurable
 Thus, if converge pointwise, is measurable.
Let be a measurable space, and let .
The following statements are equivalent:
 is measurable.
 we have .
 we have .
 we have .
 we have .
A function is measurable if
We also observe that by Proposition proposition:equivalentstatementstobeingameasurablefunction, it's sufficient to prove
so that's what we set out to do.
For and , consider the following equivalent statements:
Thus,
so
Recall that for each , the sequence is an increasing sequence in . Therefore, similarily, the following are equivalent:
Thus,
Hence,
concluding our proof!
Basically says the same as Prop. proposition:limitsofmeasurablefunctionsaremeasurable, but a bit more "concrete".
Let be a of subsets of a set , and let with be a sequence of measurable functions.
Furthermore, let
Then is a measurable function.
Simple functions
Let be a of subsets of a set .
A function is called a simple function if
 it is measurable
 only takes a finite number of values
Let be a of subsets of a set .
Let be a nonnegative measurable function.
Then there exists a sequence of simple functions such that
 for all
Converges to :
Define a function as follows. Let
and let
Then the function
obeys the required properties!
Almost everywhere and almost surely
Let be a measure or probability space.
Let be a sequence of measurable functions
 For each the function is measurable
 The function is measurable
 Thus, if the converge pointwise, then is measurable
Let be a measure space. Let be a condition in oe variable.
holds almost everywhere (a.e.) if
Let be a probability space and be a condition in one variable, then holds almost surely (a.e.) if
also denoted
Let be a complete measure space.
 If is measurable and if a.e. then is measurable.
 Being equal a.e. is an equivalence relation on measurable functions.
Convergence theorems for nonnegative functions
Problems
Clearly if with s.t. , then
hence
Therefore it's sufficient to prove that if , then there exists a nondegenerate open interval s.t. . (first I said contained in , but that is a unecessarily strong statement; if contained then what we want would hold, but what we want does not imply containment).
As we know, for every there exists such that and
Which implies
which implies
Letting , this implies that there exists an open cover s.t.
and
(this fact that this is true can be seen by considering for all and see that this would imply not being a cover of , and if , then since there exists a "smaller" cover).
Thus,
Hence, letting be s.t.
we have
as wanted!
, we have for almost every if and only if for almost every , for all .
This is equivalent to saying
if and only if
i.e. is a set of measure zero.
Then clearly
by the assumption.
Follows by the same logic:
This concludes our proof.
Integration
Notation
We let
where
Stuff
Let
where are a set of positive values.
Then the integral of over wrt. is given by
Let be a sequence of nonnegative measurable functions on . Assume that
 for each
 for each .
Then, we write pointwise.
Then is measurable, and
Let . By Proposition proposition:limitofmeasurablefunctionsismeasurable, is measurable.
Since each satisfies , we know .
 If , then since and for all we have , and .
Let and .
Step 1: Approximate by a simple function.
Let be a simple function such that and . Such an exists by definition of Lebesgue integral. Thus, there are such that , and disjoint mesurable sets such that
If any , it doesn't contribute to the integral, so we may ignore it and assume that there are no such sets.
Step 2: Find sets of large measure where the convergence is controlled.
Note that for all we have
That is, for each and ,
For and , let
And since it's easier to work with disjoint sets,
Observe that,
Then,
We don't have a "rate of convergence" on , but on we know that we are close, and so we can "control" the convergence.
Step 3: Approximate from below.
For each if , then let be such that
and otherwise, let be such that
Let , and let .
For each , and we have
Thus, , and ,
If there is a such that , then
Otherwise (if the integral is finite), then
For every and , there is an such that
For every such that
Therefore
Thus,
as wanted.
Let be any nonnegative measurable functions on .
Then
Let and observe are pointwise increasing
Properties of integrals
Let be a measure space.
If is a nonnegative measurable function, then there is an increasing sequence of simple functions such that
Given as above and for , let
and
Or a bit more explicit (and maybe a bit clearer),
For each , is a cover of . On each we have , hence on entirety of .
Consider . If , then for which in turn implies
Hence .
Finally, if , then and for all take on values
Hence, for all cases.
Furthermore, for any and , there is the nesting property
so on we have .
(This can be seen by observing that what we're really doing here is dividing the values takes on into a grid, and observing that if we're in then we're either in or ).
For , then
so again and is pointwise increasing.
Let be a measure space.
Let
 be nonnegative, measurable functions
s.t.
is defined
 be a sequence of nonnegative measurable functions.
Then
Finite sum
Scalar multiplication
Infinte sums
Let and be increasing sequence of simple functions converging to , , respectively.
Note is aslo increasing to .
By monotone convergence theorem
The argument is similar for products.
Finally, is an increasing sequence of nonnegative measurable functions, since sums of measurable functions is a measurable function.
Thus, by monotone convergence and the result for finite sums
Integrals on sets
Let be a measure or probability space.
If is a sequence of disjoint measurable sets then
Let be a measure or probability space.
If is a simple function and is a measurable set, then is a simple function.
Let be a measure or probability space.
Let be a nonnegative measurable function and .
The integral of on is defined to be
Let be a measure or probability space.
Let be a nonnegative measurable function.
If and are disjoint measurable sets, then
If are disjoint measurable sets, then
Let be a measure or probability space.
If is a nonnegative measurable function, then defined by :
is a measure on .
If , then defined by :
The (real) Gaussian measure on is defined as:
where denotes the Lebesgue measure.
A Gaussian probability measure can also be defined for an arbitrary Banach space as follows:
Then, we say is a Gaussian probability measure on if and only if is a Borel measure, i.e.
such that is a real Gaussian probability measure on for every linear functional , i.e. .
Here we have used the notation , defined
where denotes the Borel measures on .
Integrals of general functions
Let be a measure or probability space.
If is a measurable function, then the positive and negative parts are defined by
Note: and are nonnegative.
Let be a measure or probability space.
If is a measurable function, then and are measurable functions.
Let be a measure or probability space.
 A nonnegative function is defined to be integrable if it is measurable and .
 A function is defined to be integrable if it is measurable and is integrable.
For an integrable function , the integral of is defined to be
On a set , the integral is defined to be
Note that , but in the actual definition of the integral, we use .
Let be a measure or probability space.
If and are realvalued integrable functions and , then
(Scalar multiplication)
(Additive)
Let be a measure or probability space.
Let and be measurable functions s.t.
If is integrable then is integrable.
Examples
Consider with Lebesgue measure. Is integrable?
And
and
therefore
Thus, is integrable.
Lebesge dominated convergence theorem
Let be a measure or probability space.
Let be a nonnegative integrable function and let be a sequence of (not necessarily nonnegative!) measurable functions.
Asssume and all are realvalued.
If and such that
and the pointwise limit
exists.
Then
That is, if there exists a "dominating function" , then we can "move" the limit into the integral.
Since and such that , we find that and are nonnegative.
Consider
From Fatou's lemma, we have
Therefore
Consider , then
(this looks very much like Fatou's lemma, but it ain't; does not necessarily have to be nonnegative as in Fatou's lemma)
Consider
Therefore,
Which implies
Since , we then have exists and is equal to .
Examples of failure of dominated convergence
Where dominated convergence does not work
On with Lebesgue measure, consider
such that instead of as "usual" with .
Both of these are nonnegative sequences that converge to pointwise.
Notice there is no integrable dominating function for either of these sequences:
 would require a dominating function to have infinite integral, therefore no dominating integrable function exists.
 on the right, and so a dominating function would have to be above on some interval which would lead to infinite integral.
Thus, Lebesgue dominated convergence does not apply
Noncummtative limits: simple case
Noncommutative limits: another one
Consider with Lebesgue measure and
Consider and $ b > 1$ and
Note that , so is not integrable.
Consider
Commutative limits
Consider
We know that is integrable and for all and ,
By multiple applications of LDCT
Showing that in this case the limits do in fact commute.
Riemann integrable functions are measurable
All Riemann integrable functions are measurable.
For any Riemann integrable function, the Riemann integral and the Lebesgue integral are equal.
Almost everywhere and Lp spaces
If is a nonnegative, measurable function, and , then .
For , let
Observe the are disjoint and
Suppose that . This implies that on a set of positive measure, i.e.
but this implies that
Thus,
which is a contradiction, hence .
Let and be integrable.
is the set of all equivalence classes of integrable functions wrt. the equivalence relation given by a.e. equality, i.e.
If is an integrable function, the norm is
If and , the integral and norm are defined to be
If , then , and
is a real vector space with addition and scalar multiplication given pointwise almost everywhere.
Functions taking on on a set of zero measure are fine!
These functions are still the almost everywhere equal to some integrable function (even those these infinitevalued functions are integrable), hence these are in .
Let be a Cauchy sequence. Since the are integrable, we may assume we choose valued representatives.
For , let be such that for ,
and .
Thus,
and
Thus, is finite almost everywhere. Thus, this series is infinite on a set of measure zero, so we may assume the representatives are zero there and the sum is finite at each .
Thus, converges everywhere.
Let
(observe that the last part is just rewriting the ).
By monotone convergence theorem
Observe that pointwise
Applications to Probability
Notation
 is a probability space
 Random variable is a measurable function
 denotes the Borel sigmaalgebra on
 denotes the probability distribution measure for
 be a sequence of random events
 be a sequence of finitely many random events
Probability and cumulative distributions
An elementary event is an element of .
A random event is an element of
A random variable is a measurable function from to .
Let
 be a measure space and be a measurable space
 be a measurable function
Then we say that the pushforward of by is defined
The probability distribution measure of , denoted , is defined
Equivalently, it's the pushforward of by :
In certain circles not including measuretheorists (existence of such circles is trivial), you might hear talks about "probability distributions". Usually what is meant by this is for some random variable .
That is, a "distribution of " usually means that there is some probability space in which is a random variable, i.e. and the "distribution of " is the corresponding probability distribution measure!
Confusingly enough, they will often talk about " distribution of ", in which case is NOT a probability measure, but denotes a probability distribution measure of the random variable.
The cumulative distribution function of , denoted , is defined by
where is the probability distribution measure of .
The probability distribution measure is a probability measure on the Borel sets .
If is a disjoint sequence of sets in , then
so satisfies countable additivity and is a measure.
Finally,
so is a probability measure.
 is increasing
 and
is right continuous (i.e. continuous from the right)
If , then
Consider the limit as . Let
so
Then,
which, since is increasing implies
Let and . Let
The are nested, and similarily are nested.
Thus, given , there exists such that
Let so
RadonNikodym derivatives and expectations
Let
 be a rv.
 its probability distribution measure
 its cumulative distribution function
 a Borel measureable function
The following are equivalent:
 is a RadonNikodym derivative for wrt. (the Lebesgue measure but restricted to Borel measurable sets)
(2) and (3) are immediately equivalent:
iff (2) or (3) holds when considering only sets of the form .
This statement is also equivalent to (1).
Thus (1) is equivalent to (2) or (3) restricted to sets of the form .
However, sets of the form generate , so from the Carathéodory extension theorem this gives .
To prove more rigorously, let
for s.t. and none of these intervals overlap. That is all finite unions of leftclosed, rightopen, disjoint intervals.
Also let
Observe that
and that
One can show that is a premeasure space. Therefore, by the Carathéodory extension theorem, there is a measure on s.t.
Furthermore, since , is unique! But both the measures and satisfy these properties, thus
which is the definition of being a RadonNikodym derivative of wrt. Lebesgue measure restricted to the Borel σalgebra, as wanted.
A function is a probability density function for if is a RadonNikodym derivative of the probability distribution measure , wrt. Lebesgue measure restricted to Borel sets, i.e.
Expectation via distributions
Expectation of a random variable is
If is a nonnegative function that is measurable, then
If is the characterstic function, then, if ,
so
Multiplying by constants and summing over different characteristic functions, we get the result to be true for any simple function.
Given a nonnegative function , let be an increasing sequence of simple functions converging pointwise to .
Note is the increasing limit of . By two applications of Monotone Convergence
This techinque, of going from characterstic function → simple functions → general functions, is used heavily, not just in probability theory.
Independent events & BorelCantelli theorem
A collection of random events are independent events if for every finite collection of distinct indices ,
A random event occurs at if .
The probability that the event occurs is .
If are independent then are also independent.
Prove that are independent.
Consider , we want to prove
RHS can be written
which is equal to LHS above, and implies that the complement is indeed independent.
The condition that infinitively many of the events occurs at is
This is equivalent to
where we have converted the and .
Furthermore, is itself a random event.
If then probability of infinitely many of the events occuring is 0, i.e.
If the are independent and , then probability of infinitely many of the events occuring is 1, i.e.
Suppose .
Suppose are now independent and that . Fix . Then
Chebyshev's inequality
Let be a probability space.
If is a random variable with mean and variance , then
Let
Then everywhere, so
Hence,
Independent random variables
Let
 be a probability space.
A collection of σalgebras , where for all , is independent if for every collection of events s.t for all , then is a set of independent events.
A collection of random variables is independent if the collection of σalgebras they generate is independent.
A sequence of random variables is independent and identically distributed (i.i.d) if they are independent variables and for we have
where is the cumulative distribution function for .
Let and be independent.
 We have
 If or then
If and , then
Furthermore, if and , then
Consider
 first nonnegative functions
 subcase
Since is nonnegative
Thus, so .
Now consider the subcase where and .
Let and be the σalgebras generated by and .
Observe that and are measure spaces. Let be an increasing sequence of simple functions that are measurable wrt. and similarily simple increasing to and measurable.
As simple functions, these can be written as
Then,
Since increases to , by MCT
Dividing into positive & negative parts & summing gives .
Strong Law of Large numbers
Notation
are i.i.d. random variables, and we will assume
Stuff
Let be a probability space and be a sequence of i.i.d. random variables with
Then the sequence of random variables converges almost surely to , i.e.
This is equivalent to occuring with probability 0, and this is the approach we will take.
First consider .
For and , let
and
Since are i.i.d. we have
and since variance rescales quadratically,
Using Chebyshev's inequality
Observe then that with , we have
And so by BorelCantelli, since this is a sequence of independent random variables, we have
In particular, for any , there are almost surely only finitely many with
Step: showing that we can do this for any .
Consider . Observe that by countable subadditivity,
Now let , which occurs almost surely from the above. For any , let
Since , there are only finitely many s.t.
as found earlier (the parenthesis are indeed different here, compared to before). Therefore
is arbitrary, so this is true for all . Hence,
This proves that there is a subsequential limit almost surely.
Step: subsequential limit to "sequential" limit. Given , let be such that . Since are nonnegative
and therefore
and since ,
Since the first and the last expressions converge to ,t by the squeeze theorem we have
Step: Relaxing nonnegativity assumption on .
Suppose is not necessarily nonnegative. Since, by assumption, has finite expectation, is integrable. Therefore we know that the positive and negative parts of , denoted , are also integrable. Therefore we can compute the expectations
Similarily, we have that the variance of is finite, which allows us to the apply the result we found for being nonnegative to both and :
Let be the set where the mean of the positive / negative part converges. Since
(since otherwise the limit would not converge almost surely). We then have
Thus, almost surely, , and on this we have convergence, so
Concluding our proof.
Ergodic Theory
Let be a measurepreserving transformation on a measure space with , i.e. it's a probability space.
Then is ergodic if for every we have
Bochner integrable
The Bochner integral is a notion of integrability on Banach spaces, and is defined in very much the same way as integrability wrt. Lebesguemeasure.
Let be a measure space and a Banach space.
A simple function is defined similarily as before, but now taking values on a Banach space instead. That is,
with the integral
A measurable function is said to be Bochner integrable if there exists a sequence of integrable simple functions such that
where the integral on the LHS is an ordinary Lebesgue integral.
If this is the case, then the Bochner integral is defined
It can indeed be shown that a function is Bochner integrable if and only if , the Bochner space, defined similarily as L1space for functions but with the absolute value replaced by the .
Concentration inequalities
Stochastic processes
Let
 be a filtration
 be an martingale
 be an stopping time
such that one of the following holds:
 such that
and there exists a constatn s.t. for all ,
almost surely on the even that .
 such that almost surely for all
Then is a.s. welldefined and .
Furthermore, when is supersubmartingale rather than a martingale, then equality is replaced with lessgreatherthan, respectively.
Let be a supermartingale with a.s. for all .
Then for any
Let be the event that and , where we assume so that if for all .
Clearly is a stopping time and . Then by Doob's optional stopping theorem and an elementary calculation
Course: Advanced Probability
Notation
 is used as a binary operation which takes minimum of the two arguments
 is used as a binary operation which takes maxmium of the two arguments
Lecture 1
Notation
 denotes a measurable space (with a measure it becomes a measure space)
denotes the set of measurable functions wrt.
and nonnegative measurable functions
We write
Stuff
Let be a measure space.
Then there exists a unique s.t.
 for all
Linearity
for all with .

for pointwise.
There exists a unique measure on called the product measure
Let . For define
Then is . Hence, we can define
Then is and
where is the product measure.
Applying the above in both directions, we have
with
and
Conclusion:
Lecture 2: conditional expectation
Notation
 a probability space, i.e.
denotes rv, i.e. is and integrable, with expectation
Also write
or, as we're used to, instead of
Stuff
Let with .
Then
is called the conditional probability of given .
Similarily, we define
to be the conditional expectation of given .
 Quite restrictive since we require probability of to be nonzero
 Goal: improve prediction for if additional "information" is available
 "Information" is modelled by a sigmaalgebra
Let be a sequence of disjoint events, whose union is . Set
For any integrable random variable , we can define
where we set
Notice that
in (discrete) definition of conditional expectation is
Let , then
because , and each of these sets are measurable Notice that this is simply the union of intersections
which is just
But this is just since ! That is,
Which means we end up with
which is union of sets and so is random variable.
is integrable and
This is easily seen from
since and is integrable.
There's an issue with the (discrete) definition of conditional expectation though. Example:
 , and be the Lebesgue measure
Consider the case
 Then consider is a rv.
Then let
Then
Issue: if has an absolutely continuous distribution, e.g. , i.e.
then the set we're summing over is the empty set!
 This motivates the more general defintion which comes next!
Let with is a σalgebra.
A random variable is called (a version of) the conditional expectation of given by if
 is
And
So we write
 can be replaced by throughout
 If with it suffices to check for all
If with a rv., then is by condition (1) in def of conditional expectation, so it's of the form for some function ; therefore it's common to define
Let with a sigmaalgebra.
Then
 exists
 Any two versions of coincide
 Let be as in conditional expectation and let satisfy the conditions in the same def for some with almost surely.
 Let with (in because both are )
Then
since . The first equality is due to condition (2) in def of cond. expectation.
implies, by def of , that
 If , a similar argument shows that (using and )
 The reason why we did the inequality first is because we'll need that later on.
 We're going to do this by orthogonal projection in .
Assume . Since is a complete subspace of , so such has an orthogonal projection on , i.e.
Choosing for some , we get
so satisfies (1) and (2) in def of cond expectation, from equation above. But this is assuming which is not strict enough for the case when ! So we gotta do some more work.
Assume . Then and for some . By Step 1, we know that
and a.s. (by proof of (2) above). Further, let
with
which is just the set where the sequence is increasing. Then is and by MCT we get
Then, letting ,
so a.s. (and thus ) and
satisfies the condtions in def of cond expectation
For general , apply Step 2 on and to obtain and . Then
satisfies the condtions in def of cond expectation.
Let , i.e. integrable random variable, and let be a σalgebra.
We have the following properties:
 If is , then
 If is independent of , then
 If , then .
For and any integrable random variable , we have
Let be a sequence of random variables. Suppose further a.s., then a.s., for some random variable .
(conditional MCT) By MCT, we therefore have
which implies that
This is basically the conditional MCT:
(conditional Fatou's lemma)
(conditional Dominated convergence) If and for all , almost surely, for some integrable variable , then
(conditional Jensen's inequality) If is convex, then
In particular, for , we have
where we used Jensen's inequality for the inequality. Thus we have
For any σalgebra , the rv. is and satisfies, for all ,
(Tower property)
"Take out what is known": if is bounded and , then
since is then and
(actually, you need to first consider for some , and then extend to all measurable functions using simple functions as usual)
If is independent of , then
Because suppose and , then
The set of such intersections is a πsystem generating , so the desired formula follows from [SOME PROPOSITION].
From defintion of conditional expectation we know that
And since , we must also have
Let . Observe that since
and both and are , thus the their preimages must be in and
Moreover, from the definition of , we know that
which implies
i.e.
[TODO] being independent of means that
Let
since is . Then
by def of expectation. But
 Proven in proof:1.4.lecexistenceanduniquenessofconditionalexpectation
If we can show that properties (1) and (2) from def:conditionalexpectation is satisfied by , then by (2) we get LHS immediately. Measurability follows from linear combination of measurables being measurable. And observe that for all
by the fact that and are both .
Let . Then the set of random variables of the form
where is a σalgebra is uniformly integrable.
Given , we can find so that
Then choose so that
Suppose , then . In particular,
so, by Markov's inequality, we have
Then
Since our choice of was independent of , we have our proof for any σalgebra .
Martingales in discrete time
Let be a probability space.
A filtration on this space is a sequence of σalgebras such that,
We also define
Then .
A random process (in discrete time) is a sequence of random variables .
Let be a random process (discrete time).
Then we define the natural filtration of to be , given by
Then models what we know about by time .
We say is adapted to if is for all .
This is equivalent to requiring that for all .
Let
 be a filtration
is if is for each .
We say a random process is integrable if is an integrable random variable for all .
A martingale is an adapted random process (discrete time) such that
If instead
we say is a supermartingale.
And if instead
we say is a submartingale.
Every process which is martingale wrt. a given filtration is also martingale wrt. its natural filtration.
We say a random variable
is a stopping time if for all .
For a stopping time , we set
Let and be stopping times and let be an adapted process. Then
 is a stopping time
 is a σalgebra
 If , then
 is an random variable
 is adapted
 If is integrable, then is integrable.
Let
 denote a probability space
 be a filtration
 adapted to
Note that
And, since and are a stopping times,
 which also implies , since a σalgebra is closed under complements
 Similarily for
 σalgebra is closed under finite intersections and unions
Hence
Importantly this holds for all , and so
i.e. is a stopping time.
Recall
which can equivalently be written
Problem sheets
PS1
1.1
Let and let be a σalgebra. Then
First of, is because each and are and we know that linear combinations of measurables are measurable. Second,
1.2
Let
 be a nonnegative rv.
 be a version of
Then
i.e.
Further,
Course: Advanced Financial Models
Notation
Lecture 2
If is , then a.s.
Conversely, if a.s. then s.t
i.e. everything "interesting" happens in the intersection of and
If then
since is .
Suppose where .
Note tthat
where in the last equality we've used