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I have been struggling with the following for quite some time now. If anyone can give me some help, it will be much appreciated:

Let $f$ bounded, measurable and $E$ be a set of finite measure. Let $A \subset E$ be measurable. Prove that:

$\displaystyle \int_{A} f = \int_{E} f \chi_A $


marked as duplicate by user147263, user99914, Claude Leibovici, Najib Idrissi, user91500 Oct 13 '15 at 8:49

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  • $\begingroup$ You might consider the bounded measurable simple functions $<f$, and their integrals. Use this to show the equality should be a simple exercise. $\endgroup$ – awllower Jan 24 '13 at 7:25
  • 2
    $\begingroup$ I've seen this as a definition of the integral over a subset, which makes me wonder what there is to prove. $\endgroup$ – Michael Greinecker Jan 24 '13 at 7:52
  • $\begingroup$ I had also seen this as a definition in the past. It appears, however, that one can prove it from other facts, equivalent ones. $\endgroup$ – user44069 Jan 24 '13 at 7:55
  • $\begingroup$ It appears, however, that one can prove it from other facts, equivalent ones... Thus, which facts are you assuming as definitions and which ones are you not? $\endgroup$ – Did Jan 24 '13 at 8:52

This is an attempt to make it clear what needs to be shown, and hence it is not a full answer. However it was too long for a comment.

Let $(E,\mathcal{E},\mu)$ be your measure space and $A\subseteq E$ be a set from $\mathcal{E}$. Then we equip $A$ with the induced sigma-field $\mathcal{A}$, which is given by $$ \mathcal{A}:=\{A\cap E\mid E\in\mathcal{E}\} $$ and we let $\mu^A$ denote $\mu$'s restriction to $\mathcal{A}$. Then $(A,\mathcal{A},\mu^A)$ is also a measure space. If I understand your question correctly, then we want to show that $$ \int_A f_{\mid A}\,\mathrm d\mu^A=\int_E f\chi_A\,\mathrm d\mu $$ whenever $f:E\to \mathbb{R}$ is bounded and $(\mathcal{E},\mathcal{B}(\mathbb{R}))$-measurable. Here $f_{\mid A}$ denotes $f$'s restriction to $A$.

Let us furthermore assume that $f$ is non-negative. By definition of the integrals we have that $$ \int_A f_{\mid A}\,\mathrm d\mu^A=\sup\{I_{\mu^A}(s)\mid s\in\mathcal{SM}(\mathcal{A})^+,\; s\leq f_{\mid A}\}, $$ and $$ \int_E f\,\mathrm d\mu=\sup\{I_{\mu}(s)\mid s\in\mathcal{SM}(\mathcal{E})^+,\; s\leq f\}. $$ Here $\mathcal{SM}(\mathcal{E})^+$ denotes the non-negative, simple, measurable functions with respect to the sigma-field $\mathcal{E}$, and $$ I_{\mu}(s)=\sum_{j=1}^na_j\mu(A_j),\quad\text{if }\;s=\sum_{j=1}^n a_j1_{A_j}. $$ Here $s$ has a standard representation, i.e. the $A_j$'s are disjoint and their union is the whole space.

Conclusion: It suffices to show that $$ \{I_{\mu^A}(s)\mid s\in\mathcal{SM}(\mathcal{A})^+,\; s\leq f_{\mid A}\}=\{I_{\mu}(s)\mid s\in\mathcal{SM}(\mathcal{E})^+,\; s\leq f\}. $$