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I'm strugging to show the change of variable formula in Lebesgue integral. My original question is in here.

I've found a similiar question in here, and the author says $G$ is differentiable a.e, which means $f(g(t))g'(t)$ is integrable. The thing is, I have a difficulty showing this statement. Every other statement is understandable, except for the fact that $f(g(t))g'(t)$ is integrable. The whole explanation make sense under the integrability of $f(g(t))g'(t)$ I guess. But I don't think letting $g(t) = y$ and differentiating $y$ with $t$ is proper way, (my first idea).

Any other comments about the integrability of $f(g(t))g'(t)$ would be grateful. Thank you.

jason 1
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  • See comments of this answer to your second link, where a reference is given. – Anne Bauval May 13 '23 at 07:05
  • Actually I'm reading Stein's Real Analysis and solving the sane exercise I guess, which asks $f(g(t))g(t)$ is integrable (notation is slightly different), and I have a difficulty showing the integrability of the function. – jason 1 May 13 '23 at 07:12
  • I think this process is similar to my approach for proving the measurability of the same function $f(g(t))g'(t)$. Then, I guess I can deal with the integrability, except for the last step; the general function using DCT. Does the last DCT mean the limit of $\int \phi_n$ is $\int f(g(t))g'(t)dt$, where $\phi_n$ is asimple function? If that's the case, how does the DCT guarantee the integrability of $f(g(t))g'(t)$? Just the limit of finite numbers does not mean the limit is finite, as far as I know. – jason 1 May 13 '23 at 08:07
  • For reference, I used this site https://math.stackexchange.com/questions/1044021/show-that-ffxfx-is-measurable to prove the measurability, which seems similar to your reference. – jason 1 May 13 '23 at 08:15
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    I think the proposed duplicate answered your question. As for your last doubt (knowing already that $(f\circ g)g'$ is measurable) about the finiteness of $\int(|f|\circ g)g'$: approximate $|f|$ by an increasing sequence of positive simple functions $\phi_n,$ then (MCT twice) $$\int_a^b (|f|\circ g)g'=\lim\int_a^b(\phi_n\circ g)g'=\lim\int_{g(a)}^{g(b)}\phi_n=\int_{g(a)}^{g(b)}|f|<\infty.$$ – Anne Bauval May 13 '23 at 15:15
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    Ok, I see. Thank you so much. – jason 1 May 14 '23 at 05:57

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