# Bounded measurable function and integral with charcteristic function [duplicate]

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, user91500Oct 13 '15 at 8:49

• You might consider the bounded measurable simple functions $<f$, and their integrals. Use this to show the equality should be a simple exercise. – awllower Jan 24 '13 at 7:25
• I've seen this as a definition of the integral over a subset, which makes me wonder what there is to prove. – Michael Greinecker Jan 24 '13 at 7:52
• I had also seen this as a definition in the past. It appears, however, that one can prove it from other facts, equivalent ones. – user44069 Jan 24 '13 at 7:55
• 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? – Did Jan 24 '13 at 8:52

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 $$\{I_{\mu^A}(s)\mid s\in\mathcal{SM}(\mathcal{A})^+,\; s\leq f_{\mid A}\}=\{I_{\mu}(s)\mid s\in\mathcal{SM}(\mathcal{E})^+,\; s\leq f\}.$$