Let $(X,\Sigma,\mu)$ be a complete measure space. $E\subseteq X$. The textbook I'm using missed two statements that I consider true. I want to ask whether those are indeed correct and how to prove it.

  1. Let $f,~g:E\to\overline{\Bbb R}$ (extended reals) be measurable. If $f=g$ a.e and $f$ is integrable, then $g$ is integrable and $\int_E f=\int_E g$.
  2. If $E_1,~E_2,~\cdots,~E_n$ (finitely many) are disjoint measurable sets, $E=\cup_{i=1}^nE_i$ and $f:E\to\overline{\Bbb R}$ is integrable over each $E_i$($i=1$ to $n$). Then $f$ is integrable on $E$.

Yes, 1. is true. Integrability means that $\int_{\mathbb R}|f|<\infty$, and $|f|=|g|$ a.e., thus $\int_{\mathbb R}|g|=\int_{\mathbb R}|f|<\infty$ so $g$ is integrable. Now splt $f\cdot 1_E$ and $g\cdot 1_E$ into positive and negative parts and subtract.

Also 2. is true by writing $|f|$ as a sum: $|f|\cdot 1_{E_1}+\cdots +|f|\cdot 1_{E_n}$ and applying additivity of the integral.

| cite | improve this answer | |
  • $\begingroup$ Thanks. For 2. Why is $|f|\cdot 1_{E_1}$ (which has domain $E$) integrable? I know that $f|_{E_1}$ is integrable (hypothesis) and $f\cdot 1_{E_1}|_{E\setminus E_1}$ is integrable (zero function), but how to deduce that $|f|\cdot 1_{E_1}$ is integrable? $\endgroup$ – Eric Jan 5 at 7:50
  • $\begingroup$ Because $\int_{\mathbb R} |f|\cdot 1_E = \sum_{i=1}^n \int_{\mathbb R}|f|\cdot 1_{E_i}$, by additivity of the Lebesgue integral. And thus you are adding finitely many quantities each $<\infty$, so the result is also $<\infty$. Hence integrability... $\endgroup$ – pre-kidney Jan 5 at 7:53
  • 1
    $\begingroup$ Oh I answered a different question than what you asked in that last comment. I think you are asking why I can extend the domain from a smaller set to a larger set (multiplying the $1_E$) without changing integrability. This is actually just by definition of the Lebesgue integral, at least how it is normally introduced: first you define the Lebesgue integral on $\mathbb R$, then by definition the Lebesgue integral over a measurable subset $E$ of $\mathbb R$ is defined by multiplying by $1_E$. $\endgroup$ – pre-kidney Jan 5 at 7:56
  • 1
    $\begingroup$ To be fair, there are several approaches and several choices in terms of what is a definition and what is a consequence. Since your question has to do with these foundational results, which is dependent on the exact definitions used in the construction of Lebesgue integration, you should probably state which textbook you are using. $\endgroup$ – pre-kidney Jan 5 at 7:58
  • $\begingroup$ Yeah you find my confusion. Now I understand, thanks! (The book I'm using (a Chinese book by the way) start defining the integrability on a general subset of the whole space, so I didn't suddenly aware of how to show the question I asked. Since I know that other people use the definition as you mentioned and they are turn out to be equivalent, I can convinced myself it is true). Thanks for the help again! $\endgroup$ – Eric Jan 5 at 8:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.