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The following question is taken from Royden's Real Analysis $4$th edition, Chapter $4,$ exercise $10,$ page $79.$

Question: Let $f$ be a bounded measurable function on a set of finite measure $E.$ For a measurable subset $A$ of $E,$ show that $\int_A f =\int_E f \cdot \chi_A.$

My attempt:

Since $f\cdot \chi_A$ is bounded measurable on a set of finite measure $E,$ therefore it is integrable. Recall that $$\int_A f=\inf\left\{ \int_A \psi: \psi \text{ is simple and }\psi\geq f \text{ on }A \right\}$$ and $$\int_E f\cdot\chi_A = \inf\left\{ \int_E \phi: \phi\text{ is simple and }\phi\geq f\cdot\chi_A \text{ on }E \right\}.$$ For any given simple function $\psi$ such that $\psi\geq f$ on $A,$ we can extend it so that $\psi=0$ on $E\setminus A$ and this extension is still a simple function. Therefore, for any $x\in E,$ $$(f\cdot\chi_A)(x) = \begin{cases} f(x) & \text{ if }x\in A \\ 0 & \text{ if }x\in E\setminus A \end{cases} \leq \begin{cases} \psi(x) & \text{ if }x\in A \\ 0 & \text{ if }x\in E\setminus A \end{cases} = \psi(x).$$ Hence, $\int_A f \geq \int_E f\cdot\chi_A$. Similarly, let $\phi$ be a simple function such that $\phi\geq f\cdot\chi_A$ on $E.$ Then for any $x\in A,$ $$f(x) = (f\cdot\chi_A)(x)\leq \phi(x).$$ Therefore, $\int_A f \leq \int_E f\cdot\chi_A$ and hence $\int_A f = \int_E f\cdot\chi_A.$

Is my proof correct?

RRL
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Idonknow
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1 Answers1

7

Your argument that $\int_A f \geqslant \int_E f \cdot \chi_A $ is essentially correct.

To fill in some necessary detail, if $\psi \geqslant f$ on $A$, then $\psi \cdot \chi_A \geqslant f \cdot \chi_A$ on $E$ and for simple functions we have $\int_A \psi = \int_E \psi \cdot \chi_A$.

Thus,

$$\int_A \psi = \int_E \psi \cdot \chi_A \geqslant \inf_{\hat{\psi} \geqslant f \cdot \chi_A} \int_E\hat{\psi} = \int_E f \cdot \chi_A.$$

Taking the infimum of the LHS, we obtain

$$\int_A f = \inf_{\psi \geqslant f} \int_A \psi \geqslant \int_E f \cdot \chi_A.$$

To show that $\int_A f \leqslant \int_E f \cdot \chi_A$, let $\phi$ be a simple function such that $\phi \leqslant f$ on $A$. It follows that $\phi \cdot \chi_A \leqslant f \cdot \chi_A$ on $E$ and

$$\int_A \phi = \int_E \phi \cdot \chi_A \leqslant \sup_{\hat{\phi} \leqslant f \cdot \chi_A}\int_E \hat{\phi} = \int_E f \cdot \chi_A.$$

Taking the supremum of the LHS, we obtain

$$\int_A f = \sup_{\phi \leqslant f} \int_A \phi \leqslant \int_E f \cdot \chi_A.$$

RRL
  • 90,707