# Integration over a bounded set

Let $f,g: S\to \mathbb{R}$. If we assume $f$ and $g$ are integrable over $S$, then I'm trying to show:

If $f$ and $g$ agree except on a set of measure zero, then $\int_S f=\int_S g$.

Also, how can we show:

If $f(x) \le g(x)$ for $x \in S$ and $\int_S f=\int_S g$, then $f$ and $g$ agree except on a set of measure zero.

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I prefer you to use $\TeX$ syntex for mathematical formula. First show that if $f \geq 0$ and $f$ vanishes a.e., then $\int f = 0$. This can be proved by monotone convergence theorem applied to $f_n = \min(f, n)$. Now if the positiveness is discarded, write $f = f^+ - f^-$. For the seonnd one, you may use the inequality $$\int f \geq c \mu \{ f \geq c \}$$ where $c > 0$. This shows that $\mu \{ f \geq c \} = 0$ for all $c > 0$, hence $\mu \{ f > 0 \} = 0$. – Sangchul Lee Feb 16 '12 at 7:56

The first statement follows from the following fact: If $A$ is a set with measure zero, then $\int_A f\,dm=0$ for any measurable function $f$. Now, if $f=g$ on $S-A$ where $A$ is a set with measure zero, then we have $$\int_S f\,dm=\int_{S-A}f\,dm+\int_{A}f\,dm=\int_{S-A}f\,dm=\int_{S-A}g\,dm=\int_{S-A}g\,dm+\int_{A}g\,dm=\int_S g\,dm.$$
For the second statement, if $f\geq g$, then $f-g\geq 0$. For $\epsilon>0$, let $S(\epsilon)=\{x | f(x)-g(x)\geq \epsilon\}$ which is a measurable set. Then we have $$\tag{1}\int_{S(\epsilon)}(f-g)\geq\epsilon\,m(S(\epsilon)).$$ By assumption, we have $$\tag{2}0=\int_{S}(f-g)dm=\int_{S-S(\epsilon)}(f-g)dm+\int_{S(\epsilon)}(f-g)dm\geq \int_{S(\epsilon)}(f-g)dm$$ since $f-g\geq 0$ on $S$. Combining $(1)$ and $(2)$, we have $$m(S(\epsilon))=0\mbox{ for all }\epsilon>0.$$ Note that $$\{x\in S| f(x)\neq g(x)\}=\{x\in S| f(x)>g(x)\}=\bigcup_{n=1}^\infty S(\frac{1}{n}).$$ This implies that $\{x\in S| f(x)\neq g(x)\}$ has measure zero.