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Let $f:\mathbb R\to\mathbb R$ be a non-negative Lebesgue measurable function such that $\int_I f(x)dx<\infty$ for all finite intervals $I=[a,b]$. For every Lebesgue-measurable set $A\subset\mathbb R$, define $$ \mu(A)=\int_Af(x)dx. $$ Prove that if $g:\mathbb R\to\mathbb R$ is a Lebesgue-measurable function, then $$ \int_{\mathbb R} g(x)d\mu(x)=\int_{\mathbb R} g(x)f(x)dx. $$

I know that this is a direct result of Radon–Nikodym theorem but can we show the identity above only with the help of basic Lebesgue-integral theory? WLG we can assume $g(x)\ge 0$ and consider the simple functions approximation. However, I find it hard if we only use the definition of the integral of non-negative functions. Can someone give me some hints? Thank you.

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  • $\begingroup$ What counts as "basic Lebesgue-integral theory"? E.g. is the monotone convergence theorem allowed? $\endgroup$ Commented Jun 16, 2019 at 16:20
  • $\begingroup$ @carmichael561 Sorry, I didn't make this clear enough. Yes, the monotone convergence thm, dominated convergence thm, Fatou's lemma are counted. $\endgroup$ Commented Jun 16, 2019 at 16:22
  • $\begingroup$ Have you tried proving this statement for $g$ defined on finite intervals? $\endgroup$ Commented Jun 16, 2019 at 16:24
  • $\begingroup$ @SeanNemetz I did not... should have realized this earlier... $\endgroup$ Commented Jun 16, 2019 at 16:36

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The desired equality holds when $g$ is an indicator function by definition, hence when $g$ is a simple function by linearity. And if $g\geq 0$, let $\{\phi_n\}$ be a sequence of simple functions increasing to $g$, then by two applications of the monotone convergence theorem we have $$ \int g(x)\;d\mu(x)=\lim_{n\to\infty}\int \phi_n(x)\;d\mu(x)=\lim_{n\to\infty}\int \phi_n(x)f(x)\;dx=\int g(x)f(x)\;dx$$

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    $\begingroup$ Thank you! I should have considered first when $g$ is an indicator function... $\endgroup$ Commented Jun 16, 2019 at 16:34

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