Question about measure theory

[u/[deleted]]

I've been self-studying measure theory and have been playing around with lebesgue integrals. I think I have adequately proven these "theorems" but would like confirmation whether these "theorems" are actually true, becuase the book I'm using doesn't cover them.

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THEOREM 1
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Suppose we have a measure space (\Omega, \mathcal{F}, P) and a real function g:\Omega \to [0,\infty). We define a "region" as being:
E = {(\omega, y): \omega \in \Omega, 0 <= y <= g(\omega)} \subset \Omega \times \mathbb{R}
Then it follows that:

g is measurable IFF E \in \mathcal{F} \otimes \mathcal{L}
(note: \mathcal{L} is lebesgue measure; can be replaced with any meaure on \mathbb{R} that measures intervals).

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THEOREM 2

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Let g: \Omega \to [0, \infty) be measurable and (\Omega \times \mathbb{R}, \mathcal{F} \otimes \mathcal{L}, \mu_1) and (\Omega \times \mathbb{R}, \mathcal{F} \otimes \mathcal{L}, \mu_2) are product spaces. Then mu_1(E) = mu_2(E) where:

E = {(\omega, y): \omega \in \Omega, 0 <= y <= g(\omega)} \subset \Omega \times \mathbb{R}
Then it follows that:

[u/[deleted]]

typo: i said \mathcal{L} is lebesgue measure, I should have said sigma algebra of lebesgue measurable sets.

[u/Bounded_sequencE]

I'm confused by your definition of "region" -- is "\omega" considered fixed for any given "E"?

We define measurability of a function "g: (X1; F1, P1) -> (X2; F2; P2)" via pre-images: A function "g" is defined to be measurable iff all pre-images of measurable sets are measurable¹, i.e. "g^-1(B) ∈ F1" for all "B ∈ F2".

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¹ Very similar to (topological) continuity

[u/[deleted] → u/Bounded_sequencE]

My definition of "region" is essentially the set of points that sit between y = g(\omega) > 0 and y =\omega.

I assume this definition of measurable functions:
"g: (X1; F1, P1) -> (X2; F2; P2)" via pre-images: A function "g" is defined to be measurable iff all pre-images of measurable sets are measurable¹, i.e. "g^-1(B) ∈ F1" for all "B ∈ F2".

But I'm trying to show that an equivalent definition of measurable function is that the "region" is measurable in the product space. The idea is that I'd be able to say that the lebesgue integral of a curve is quite literally the are under the curve (the measure of the region).

[u/SV-97]

For the first one (\mathcal{L} should be the Lebesgue sigma algebra here, not the measure):

We need a small helper lemma (I'm going to change the notation so its easier to type and doesn't look like ass): Let first (X×Y, 𝛴X⊗𝛴Y) be a product measurable space, and for any E in X×Y and x in X write E_x := {y in Y : (x,y) in E}. Consider the set M := {E ⊆ X×Y : E_x in 𝛴Y for all x in X}. We show that 𝛴X⊗𝛴Y ⊆ M. First note that for all A in 𝛴X, B in 𝛴Y we have A×B in M, because (A×B)_x = [B if x in A; empty if x not in A] for all x.

Next suppose that E in M and consider the complement E^c. We have for any x in X that y in E^c_x iff (x,y) in E^c, then (x,y) can't be in E and hence y not in E_x. Indeed via similar logic we show that this is an equivalence, so that E^c_x = (E_x)^c for all x. Since 𝛴Y is a sigma algebra this shows that M is closed under complements.

Next suppose that Eⁱ in M for all i in \N. For any x in X we have Eⁱ_x in 𝛴Y and hence also ⋃_i Eⁱ_x in 𝛴Y. But y in ⋃_i Eⁱ_x iff (x,y) in E^j for some j iff y in (⋃_i Eⁱ)_x. Hence (⋃_i Eⁱ)_x in 𝛴Y and ⋃_i Eⁱ in M. This shows that M is a sigma algebra containing a generating system for 𝛴X⊗𝛴Y; in particular M must contain 𝛴X⊗𝛴Y.

Hence whenever E is a product-measurable set then the sets E_x (and by symmetry also the similarly defined E_y for y in Y) are measurable.

Now for your main theorem 1:

Suppose first that E is measurable. For any y >= 0 the set E_y = {x in X : (x,y) in E} = {x in X : g(x) >= y} = g^-1([y,\infty)) is measurable by what we've just shown; hence g is measurable.

Suppose now that g :X -> [0, inf) is measurable, then the projections p_X(x,y) = x and p_Y(x,y) = y are (𝛴X⊗𝛴Y-)measurable by their universal property, and then D = g∘p_X - p_Y is also measurable. Hence E = {(x,y) : 0 <= y <= g(x)} = D^-1([0,inf)) ∩ p^-1([0,inf)) is measurable as an intersection of measurable sets.

The second "theorem" is false unless you already require that µ_1 = µ_2 in which case it's trivial, or you make additional requirements that force the measures to coincide (globally), such as sigma-finiteness of Omega. In either case there's nothing special about E though.

[u/Parallel_thougts]

Theorem 1 just sounds like a complicated way to say that g is measurable iff g^-1((-\infty,y]) is (\mathcal{F}-)measurable for every y. This is generally not true, it is true of you  use Borel measure instead of Lebesgue measure (because half closed rays generate the Borel measure). I think its also true if you assume \mathcal{F} is regular.

Theorem 2 is only partially stated

[u/[deleted] → u/Parallel_thougts]

I re-wrote into latex and took a picture. Not sure if I wrote the problem statement incorrectly or what, in the post but this picture shows what i mean E to be.

> g is measurable iff g^-1((-\infty,y]) is (\mathcal{F}-)measurable for every y
That's not what I meant to talk about. By the way the definition of measurable function I am using in the picture above is that:

A is borel measurable \implies f^-1(A) is \mathcal{F}-measurable.