We have $$ \mathbf{E}[g(X)] \stackrel{\text{df}}{=} \int_{\Omega} g(X) ~ \mathrm{d}{P} = \int_{\Bbb{R}^{k}} g ~ \mathrm{d}{P_{X}}, $$ where $ X: \Omega \to \Bbb{R}^{k} $ is a continuous random vector with pdf $ f_{X}: \Bbb{R}^{k} \to \Bbb{R} $ and $ g: \Bbb{R}^{k} \to \Bbb{R} $ is any function such that $ \mathbf{E}[g(X)] $ exists. Additionally, we know that $$ \mathbf{E}[g(X)] = \int_{\Omega} g ~ \mathrm{d}{P_{X}} = \int_{\Bbb{R}^{k}} g f_{X} ~ \mathrm{d}{\mu}, $$ where $ \mu $ is the Lebesgue measure on $ \Bbb{R}^{k} $.

My question is on the following step: $$ \mathbf{E}[g(X)] = \int_{\Bbb{R}^{k}} g f_{X} ~ \mathrm{d}{\mu} = \int_{\Bbb{R}} \cdots \int_{\Bbb{R}} g(x_{1},\ldots,x_{k}) {f_{X}}(x_{1},\ldots,x_{k}) ~ \mathrm{d}{x_{1}} \cdots \mathrm{d}{x_{k}}, $$ where the second equality follows because here the Lebesgue integral is equal to the Riemann integral. I don’t see why this is so.

The only time I know for sure that the integrals are equal is when we are talking about a (bounded) Riemann-integrable function on a bounded domain in $ \Bbb{R}^{k} $. Here, we don’t have such a domain, but I guess that since we know that the function $ g $ is such that $ \mathbf{E}[g(X)] $ exists, perhaps that lets us know that the Lebesgue and Riemann integrals are equal?


  • $\begingroup$ You’re right. Firstly, $ \Omega \subseteq \Bbb{R}^{k} $ may not be bounded, and although $ g \in L^{1} \! \left( \Bbb{R}^{k} \right) $, it may not be almost-everywhere continuous. $\endgroup$ – Berrick Caleb Fillmore Nov 30 '15 at 19:07

The Lebesgue integral and the proper Riemann integral always coincide provided the latter exists. Additionally, the Lebesgue integral and the improper Riemann integral also coincide provided they both exist. This follows from the monotone convergence theorem: given $f$ satisfying the conditions and an increasing sequence of compact sets $K_n$ whose union is the whole space, you have

$$\int_{\bigcup_{n=1}^\infty K_n} f^+ d \mu = \lim_{n \to \infty} \int_{K_n} f^+ d \mu$$

and the same for $f^-$, because of the monotone convergence theorem. But you also have equality with the proper Riemann integrals. So the integrals for $f^+$ and $f^-$ separately agree. Now you subtract and you get the equality.

Note that both conditions are separately necessary. That is because it is possible for the improper Riemann integral to exist only in the sense of conditional convergence, whereas the Lebesgue integral can only exist in the sense of absolute convergence.

  • $\begingroup$ Sorry, I am not quite sure what you are showing. You are showing that the lebesgue and improper Riemann integral coincide provided they both exist? If so, you are doing this by using the MCT to find $\int f^+ d\mu$, and $\int f^- d\mu$. If I'm not wrong yet, then I get lost at this next step. For every $K_n$ we know $\int_{K_n} f^+ d\mu$ is equivalent to the corresponding proper reman integral. Then I don't know. We subtract something? $f^+ - f^-$ perhaps? to get $f$? $\endgroup$ – majmun Nov 30 '15 at 19:27
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    $\begingroup$ @majmun For every $K_n$ the Lebesgue integral over $K_n$ is equivalent to the proper Riemann integral over $K_n$. When you take $n \to \infty$, the monotone convergence theorem tells you that you get the Lebesgue integral over the whole space. Also, taking that limit is the definition of the improper Riemann integral. So the Lebesgue integral and improper Riemann integral agree for $f^+$ and for $f^-$ and therefore for their difference. $\endgroup$ – Ian Nov 30 '15 at 19:37
  • $\begingroup$ Wow. That seems like a very powerful result. Thank you very much, especially for the clarification. $\endgroup$ – majmun Nov 30 '15 at 19:42
  • $\begingroup$ @majmun It's satisfying, but not very powerful, really; it essentially says "if there are multiple ways to define the integral, the values are the same". (It is also true with en.wikipedia.org/wiki/Henstock%E2%80%93Kurzweil_integral I think.) If you were said this to a scientist or an engineer and they understood you, they would probably respond with "well yeah, if that didn't happen then you'd need to redefine everything!" $\endgroup$ – Ian Nov 30 '15 at 19:55

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