2

I've been introduced to the Fubini-Tonelli theorems in my probability/measure theory class in the following way: Fubini-Tonelli Theorems

or if you want a similar looking pdf: https://www.cmi.ac.in/~prateek/measure_theory/2010-10-06.pdf. Notice that all the integrals in the image and the pdf are w.r.t. measures like $\mu, \nu$ and NOT variables like $t,x,y$.

Question setup: Now, looking at the answer to the following question: Intuitive explanation for $\mathbb{E}X= \int_0^\infty 1-F(x) \, dx$, one of the integrals is $dP$ (a probability measure) and one is $dt$, which is not a measure. This makes me uncomfortable because then it doesn't "look like" we can use the above versions of Fubini/Tonelli on it, because the connotation of $dt$ is Riemann integration and not Lebesgue integration.

Question: can we change the $dt$ integral to be something like $d\lambda$ (where $\lambda$ denotes Lebesgue measure)? When are we sure we can interchange between $dt$ (or $dx$ or $dy$) integrals with integrals w.r.t measures like $\lambda$, especially in these infinite cases?

The reason I specify the infinite case (and the one in the linked question in particular, which is special because the inside function, $f(x) = P(X<x)$, is a monotonically decreasing bounded function) is because I know for the bounded, interval case, we have: General condition that Riemann and Lebesgue integrals are the same.

Basically, can someone tell me how to get the $dt,dx,dy$ integrals into the measure theoretic form, and the criterion for knowing when I can do this interchange or not?

I get that this may be a simple misunderstanding of notation on my part, so if that's the case I hope someone can explain the correct way of interpreting this notation.

Community
  • 1
D.R.
  • 8,691
  • 4
  • 22
  • 52
  • 1
    The $dt$ in your Question setup paragraph is nothing but integration w.r.t. Lebesgue measure. The proof of the formula $EX =\int_0^{\infty} P(X>t) dt$ is completely in the framework of Fubini/Tonelli Theorem – Kavi Rama Murthy Nov 20 '19 at 08:49
  • @KaboMurphy It looks a lot like normal Riemann integration to me... I guess I just don't understand how to interpret the symbols $\int_0^\infty f(t) dt$. It doesn't look like it's involving a measure at all, so if it actually is, I want to know how we can say that it is. Why is it justified we can just encode BOTH Riemann and Lebesgue integration into one "$dt$"? And I know in general the Riemann and Lebesgue integrals are different, so I want to be careful. – D.R. Nov 20 '19 at 08:53
  • Doesn't the proof you linked still go through with the Lebesgue measure producted with the probability measure? – Mark Nov 20 '19 at 08:53
  • generally, working with measures, we omit the writing of specific variables, that is, it is generally written as $,\mathrm d \mu $ instead of $,\mathrm d \mu (x)$ or $,\mathrm d x$, however all these expressions have the same meaning in context. By the other hand the expression $\int_0^\infty f(x) d x$ is an improper integral of Riemann, however if $f$ is absolutely integrable (that is, if $\int_0^\infty |f(x)| dx$ exists as a finite value) then $\int_0^\infty f(x) dx=\int_{(0,\infty)}f d\lambda$ where $\lambda$ is the Lebesgue measure – Masacroso Nov 20 '19 at 09:07
  • That's a very involved way of introducing the theorem of Fubini-Tonelli. Actually, that theorem is very natural; just like you can swap two finite sums, such as $$\sum_{i=1}^n \sum_{j=1}^m a_{ij}= \sum_{j=1}^m \sum_{i=1}^n a_{ij}, $$ simply because the sum is commutative, you can swap two integrals, under minimal assumptions. That's all. Measurability, sigma-algebras, etc... all of that is technical blah-blah which comes much later. – Giuseppe Negro Dec 09 '19 at 08:25

1 Answers1

1

All that you need to know are these theorems, that I will state here without proof:

Theorem 1: every Riemann integrable function is Lebesgue integrable, that is, if $\int_a^b f(x)\,\mathrm d x$ exists for some pair $-\infty <a<b<\infty $, then $\int_a^b f(x) \,\mathrm d x=\int_{[a,b]}f\,\mathrm d \lambda $ where the latter is an integral of Lebesgue respect to the Lebesgue measure $\lambda $.

Theorem 2: if $f$ is Lebesgue integrable (w.r.t. the Lebesgue measure in the real line) and the set of discontinuities of $f$ have measure zero then $f$ is improperly Riemann integrable and $\int_{\Bbb R }f\,\mathrm d \lambda =\int_{-\infty}^\infty f(x)\,\mathrm d x$.

Theorem 3: let $f:\Bbb R \to \Bbb R $ Riemann integrable in any bounded interval. Then if $f$ is non-negative or the improper integral of Riemann $\int_a^b|f| \,\mathrm d x$ is finite (for $-\infty\leqslant a\leqslant b\leqslant \infty$) then $\int_{a}^{b}f(x)\,\mathrm d x=\int_{(a,b)}f\,\mathrm d \lambda $, where the first is an improper integral of Riemann, and the latter is an integral of Lebesgue w.r.t. the Lebesgue measure.

Almost any other case of functions $f:I\to \Bbb R $, for some interval $I\subset \Bbb R $, can be reduced to one of the theorems above considering the trivial extension $\tilde f:\Bbb R \to \Bbb R$ defined by $$ \tilde f(x):=\begin{cases} f(x),&x\in I\\ 0,&\text{ otherwise } \end{cases} $$ Hope it helps.

Community
  • 1
Masacroso
  • 30,417
  • Is the following case included in Theorem 3: $\int_{-\infty}^\infty |f(x)| dx = \infty \iff \int_{\mathbb R} |f| d\lambda = \infty$? – D.R. Nov 20 '19 at 16:24
  • @D.R. yes, because the function $|f|$ is non-negative. I edited a bit the last theorem to make it more general or easier to apply – Masacroso Nov 21 '19 at 02:21
  • Thank you. Where might I find proofs of these 3 theorems? – D.R. Nov 21 '19 at 02:27
  • 1
    @D.R. I studied it in the textbook Analysis III of Amann and Escher, however this textbook is not easy to get into. The definition of the integrals and the monotone convergence theorem make easy to prove theorem 1 and 3. The theorem 2 is more involved and you can see a proof here. However I extended theorem 2 to the case of improper integrals of Riemann assuming that $f$ is Lebesgue integrable, what is also easy to prove using the definition of improper integral of Riemann – Masacroso Nov 21 '19 at 02:45