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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. –  Harald Hanche-Olsen Nov 11 '12 at 6:43
    
@HaraldHanche-Olsen My book is Royden's Real Analysis 4e. Is there a less big cannon way of doing this? –  emka Nov 11 '12 at 6:46
    
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. –  Harald Hanche-Olsen Nov 11 '12 at 7:16
    
How did you demonstrated in the BCT that the convergence could be a.e.??? –  André Caldas Nov 14 '12 at 16:19
    
Why does $\phi_n$ have to be simple? Can't you simply take $\phi_n = f$ and use the bounded convergence theorem? –  André Caldas Nov 14 '12 at 16:21

1 Answer 1

up vote 2 down vote accepted

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 $h$. –  emka Nov 11 '12 at 10:44
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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. –  commenter Nov 11 '12 at 11:07

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