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Given the following: $f$ is a bounded measurable function on a set of finite measure $E$. Assume $g$ is also bounded and $f=g$ a.e. on $E$. Show that $\int_E f = \int_E g$.

The most "powerful" tool right now is the bounded convergence theorem. So I'm proceeding from there.

WLOG suppose $f,g \geq 0$. Since $f \geq 0$, there exists a sequence of simple functions $\phi_n$ such that $\phi_n \leq f$ and $\phi_n \rightarrow f$. So by the Bounded Convergence Theorem, $\lim\limits_{n \rightarrow \infty} \int_E \phi_n = \int_E f$.

Because $f = g $ a.e. on $E$, then $\lim\limits_{n \rightarrow \infty} \int_E \phi_n = \int_E g$.

Is this a proper application of the BCT?

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    Minor nit: I think you mean *simple* functions $\phi_n$, not *special* functions. Your proof seems correct, but I am left with a nagging feeling that you are dragging out a much too big cannon with which to shoot this rather small bird. Exactly what is the best proof, though, depends a lot on details of how the theory is developed.2012-11-11
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    @HaraldHanche-Olsen My book is Royden's Real Analysis 4e. Is there a less big cannon way of doing this?2012-11-11
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    I did learn this from Royden myself, but a much earlier version. And it's been too long for me to remember how he did it. But I think it makes more sense to establish that $\inf f\le \int g$ if $f\le g$ a.e. Now assuming that the additivity of the integral has been established, what you want can be fairly easily reduced to showing that $\int f=0$ if $f\ge0$ (everywhere) and $f=0$ a.e. And that, in its turn, follows directly from the definition of the integral, no big theorems required.2012-11-11
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    How did you demonstrated in the BCT that the convergence could be a.e.???2012-11-14
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    Why does $\phi_n$ have to be simple? Can't you simply take $\phi_n = f$ and use the bounded convergence theorem?2012-11-14
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    What is the bounded convergence theorem (BCT)?2018-12-10

2 Answers 2

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As Harald Hanche-Olsen said, your proof is perfectly correct and I like his second solution.

Here's yet another approach, using the inequality $\lvert \int f\rvert \leq \int \lvert f\rvert$ which follows directly from the definition of the integral as a supremum over simple functions and $|f| = f^+ + f^-$ where $f = f^+ - f^-$ is the decomposition of $f$ into positive and negative parts.

  1. If $h$ is bounded, measurable and $h = 0$ a.e. on $E$ then $\int_E h = 0$: there exists a null set $N \subset E$ such that $h = 0$ on $E \setminus N$. Then $$ 0 \leq \left\lvert \int_E h \right\rvert \leq \int_N \underbrace{\lvert h\rvert}_{\leq C} + \int_{E \setminus N}\underbrace{\lvert h\rvert}_{=0} \leq C \mu(N) = 0 $$
  2. Apply 1. to $h = f-g$.
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    The only thing I don't follow is why $h$ is not decomposed into $h^+$ and $h^-$. Instead we just call each of those $h$.2012-11-11
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    @emka: In 1. Jon suppresses the middle step $$\left\lvert \int_E h \right\rvert \leq \int_E |h| = \int_{N} |h| + \int_{E\setminus N}|h|$$ where he's using the inequality he mentioned in the paragraph before it.2012-11-11
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    @Jon how can we say that set $E \setminus N$ is measurable here?what if $E \setminus N$ is nonmeasurable and $N$ has measure zero,but $E$ is measurable?2016-02-19
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    If $E$ and $N$ are measurable, then so is $E \setminus N = E \cap N^C$.2018-10-13
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Let $A\subset E$ be the set of measure zero on which $f\ne g$. Then $A\cup E\setminus A=E$ yields that $E\setminus A$ is measurable, as it is the complement in a measurable set of a set of measure zero. Note also that $g$ is measurable on $E$. Where $f=g$ this is clear. Let $M\in \mathbb{R}$ then $$ \{ x: g(x)>M\}\cap A\subseteq A $$ a set of measure zero. But subsets of measure zero sets have measure zero, and thus are measurable. Finally, since both $A$ and $E\setminus A$ are measurable, so is $f$ on the sets, since \begin{align*} &\{ x:\;x\in A, \;f(x)>m\}=\{ x:f(x)>m\}\cap A\\ &\{ x:\;x\in E\setminus A, \;f(x)>m\}=\{ x:f(x)>m\}\cap E\setminus A \end{align*} the intersection of measurable sets. Thus, \begin{align*} &\int_{E\setminus A}f=\int_{E\setminus A}g\\ &\stackrel{m(A)=0\;\text{and} \; g\;\text{bounded}}{\implies} \int_{E\setminus A}f+\int_{A}f=\int_{E\setminus A}g+\int_{A}g\\ &\implies \int_Ef=\int_Eg \end{align*} With the final implication following since $$ A\cap B=\emptyset\implies \int_{A\cup B}f=\int_Af+\int_Bf $$ for $A,B$ measurable subsets of a set of finite measure and $f$ bounded.