The question was:(From Royden "Real Analysis" fourth edition)

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 proof was:

Let $f$ be a bounded measurable function on a set of finite measure $E.$ Then by Theorem 4 on page 74, $f$ is integrable over $E.$

Now, by The definition on page 73, $f$ is Lebesgue integrable over $A$ (because $A$ is a measurable subset of $E$ by the assumption of the problem and hence has finite measure) if and only if the following holds: $$\begin{align*}\int_A f &= \sup\{\int_A\varphi : \varphi \text{ is simple and } \varphi \leq f\} \\ &= \inf\{\int_A\psi: \psi \text{ is simple and } f \leq \psi\}.\end{align*}$$

Also, $f\cdot \chi_A$ is integrable if and only if the the following holds: $$\begin{align*}\int_E f \cdot \chi_A &= \sup\{\int_E\varphi : \varphi \text{ is simple and } \varphi \leq f\cdot \chi_A\} \\ &= \inf\{\int_E\psi: \psi \text{ is simple and } f \cdot \chi_A \leq \psi\}.\end{align*}$$

Now, since $\int_A f=\inf\{ \int_A \psi: \psi \text{ is simple and } \psi\geq f \text{ on }A\}$ and $\int_E f\cdot\chi_A = \inf \{ \int_E \phi: \phi \text{ is simple and }\phi\geq f\cdot\chi_A \text{ on }E \}.$

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 $x\in E,$ $$(f \cdot \chi_A)(x) = \begin{cases} f(x) & \text{ if } x\in A \\ 0 & \text{ if } x\in E\setminus A \end{cases} \leq \begin{cases} \psi(x) & \text{ if }x\in A \\ 0 & \text{ if }x\in E\setminus A \end{cases} = \hat{\psi}(x).$$

Now, if $\psi \geq f$ on $A$, then $\psi \cdot \chi_A \geq f \cdot \chi_A$ on $E$ by monotonicity of integration proposition 2 or Theorem 5 and because for simple functions we have $\int_A \psi = \int_E \psi \cdot \chi_A$.


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

Taking the infimum of the LHS, we obtain

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

Hence, $\int_A f \geq \int_E f\cdot\chi_A$.

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

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

Taking the supremum of the LHS, we obtain

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

But there were some comments I received on my solution :

1-Why is $f$ measurable on $A$?

2-Why is $f\cdot \chi_{A}$ measurable?

3- Prove that for simple functions we have $\int_{A} \psi = \int_{E} \psi \cdot \chi_{A}$?

Could anyone help me in answering those comments please?

Note: we are not allowed to use any material from the book after page 79.

  1. Let $M \subseteq \mathbb{R}$ be Borel measurable. Since $f$ is a measurable function, the preimage $f^{-1}(M)$ is measurable. Since $A$ is measurable, $f^{-1}(M) \cap A$ is measurable.
  2. As before, let $M \subseteq \mathbb{R}$ be Borel measurable. Then $$(f\cdot \chi_A)^{-1}(M) = \begin{cases}f^{-1}(M) \cap A, & \text{ if } 0\not\in M \\ (f^{-1}(M) \cap A) \cup (E\setminus A), & \text{ if } 0 \in M \end{cases}$$ which is measurable in either case since $f$ is a measurable function and $A \subseteq E$ is measurable.
  3. Let $N \subseteq E$ be measurable. Then $$\int_A \chi_N = |N\cap A| = \int_E\chi_{N\cap A} = \int_E\chi_N\cdot \chi_A$$ shows that $\int_A\psi = \int_E \psi\cdot \chi_A$ is true when $\psi = \chi_N.$ By linearity of the integral, it is also true when $\psi$ is a simple function.
  • $\begingroup$ You should probably clarify that you mean that $M$ is Borel measurable. For a Lebesgue measurable function, the inverse image of a Lebesgue measurable set need not be Lebesgue measurable (see math.stackexchange.com/questions/479441/…). $\endgroup$ – cmk Oct 16 at 22:44
  • $\begingroup$ @cmk In the present context, $M$ must be measurable in the codomain of $f,$ which certainly means $M$ is implicitly Borel measurable for Lebesgue measurable $f.$ In any case, thanks for pointing out the possible confusion someone might have when learning this topic. $\endgroup$ – Brian Moehring Oct 16 at 23:09
  • $\begingroup$ For 3, what do you mean by the absolute value sign in $|N \cap A|$? $\endgroup$ – Emptymind Oct 17 at 12:04
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    $\begingroup$ @Intuition The measure of the the set $X$ is sometimes written $|X|.$ In particular, in this type of case where have no "name" for the measure (common names being $\nu,$ $\mu$ or $m$), $|N \cap A|$ allows us to denote the measure of $N \cap A$ without introducing another symbol. $\endgroup$ – Brian Moehring Oct 18 at 3:43

$1$. If $V$ is an arbitrary open set in $\mathbb{R}$ and $f|_A$ denotes the restriction of $f$ to $A$, then $(f|_{A})^{-1}(V)=f^{-1}(V)\cap A.$ A real-valued function is Lebesgue measurable if and only if its inverse image of an open set is measurable. Since $f$ is measurable, so is $f^{-1}(V),$ and $A$ is measurable by assumption. So, their intersection is measurable.

$2$. For any measurable functions, $f$ and $g$, I claim that $fg$ is measurable. First, note that $$fg=\frac{(f+g)^2-f^2-g^2}{2},$$ so it will suffice to show that if $h$ is measurable, then so is $h^2$ (since the sum of measurable functions is measurable and so is a measurable function times a constant, both of which I assume you know; if not, they follow from the composition property that I'll cite below). Note that this is a composition of $h$, which is measurable, and $x^2$, which is continuous, so their composition will be measurable. This is because if $u$ is continuous and $v$ is measurable, then $u\circ v$ is measurable, too; this follows from $(u\circ v)^{-1}(V)=v^{-1}\circ u^{-1}(V)$, since $u^{-1}(V)$ is open for $V$ open by continuity and $v$ is measurable, so the inverse image of an open set is measurable. If you don't like using a result like this, then you can instead check measurability on $(a,\infty),$ for any $a$. The inverse image for $a<0$ is everything, and for $a\geq 0$ is $$\{x: h^2(x)>a\}=\{x:h(x)>\sqrt{a}\}\cup\{x:-h(x)>\sqrt{a}\},$$ which is clearly measurable.

In any case, $f$ and $\chi_A$ are measurable, so is their product. You can do this more explicitly since you're working with something like a characteristic function, but we can pretty easily work in more generality, as shown.

$3$. Let $\psi(x)=\sum\limits_{j=1}^nc_j\chi_{A_j}(x),$ where $A_j$ are disjoint and measurable. Then, $$\int\limits_A \psi=\sum\limits_{j=1}^nc_jm(A_j\cap A),$$ and \begin{align*}\int\limits_E \psi\chi_A=\int\limits_E \sum\limits_{j=1}^nc_j\chi_{A_j}\chi_A&=\int\limits_E \sum\limits_{j=1}^nc_j\chi_{A_j\cap A}=\sum\limits_{j=1}^n c_j\int\limits_E \chi_{A_j\cap A}\\ &=\sum\limits_{j=1}^nc_jm(A_j\cap A). \end{align*} So, they indeed match. Here, I used the definition of the integral of a simple function, properties of characteristic functions (what their product looks like), and linearity of the integral.

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    $\begingroup$ @Intuition it suffices to check measurability on open sets (that is, to show that the inverse image of any open set is measurable). $\endgroup$ – cmk Oct 16 at 22:17
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    $\begingroup$ This is because open sets generate the Borel sets. $\endgroup$ – cmk Oct 16 at 22:23
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    $\begingroup$ @Intuition the question was to show that $f$ is measurable on $A$. That means that $f|_A:A\rightarrow\mathbb{R}$ is a Lebesgue measurable function. To show this, we need to show that $(f|_A)^{-1}(V)$ is Lebesgue measurable for every open set $V\subset\mathbb{R}.$ I don't need to have an open $V$ or anything. I need to show that the statement is true for an arbitrary open $V$. $\endgroup$ – cmk Oct 16 at 22:54
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    $\begingroup$ @Intuition whoops, typo. Fixed now. Thanks! $\endgroup$ – cmk Oct 17 at 0:46
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    $\begingroup$ @Intuition for my representation of $\psi,$ yes, I'm using pages $71-72$. The reason that I'm intersecting with $A$ when I'm taking the measure is because the $A_j$'s are measurable subsets of $E$, and I'm integrating over just $A$ (I should've noted that $A_j\subset E$ for each $j$). It's pretty easy to see that $\int_A\chi_{A_j}=m(A\cap A_j)$, then we just use linearity. Remember, $\chi_{A_j}$ is only non-zero on $A_j$, and I'm integrating over $A$. So, when we calculate the integral, we're only going to pick up the measure of the part of $A_j$ that's contained in $A$. $\endgroup$ – cmk Oct 17 at 13:15

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