About two functions whose Lebesgue integral on all sets of a $\sigma-$algebra are equal

Question

Let $X$ be an infinite set and $\mathcal{F}$ be a $\sigma-$algrbra with infinite sets on $X$. Given on $X$ a measure $\mu$.

Let $f$ and $g$ be two $\mathcal{F}-$measurable functions. Is it necessary that if $$ \int_A f d\mu = \int_A g d\mu, \forall A \in \mathcal{F}$$ then $f = g$ ($\mu-$a.e)?

I feel that the answer for this is positive, but I can't prove the statement above nor give a counter-example. Please give me a hint. Thank you.

Answer 1

Yes. Assume $f$ and $g$ are $\mathcal{F}$-measurable. Then by assumption you have

$$0=\int_{\{f>g\}}fd\mu-\int_{\{f>g\}}gd\mu=\int_{\{f>g\}}(f-g)d\mu$$

Since $(f-g)\cdot1_{\{f>g\}}\ge 0$, you have $\mu(\{f>g\})=0$ and $\mu(\{f<g\})=0$ analogously.

Remark: This property is used to define conditional expectation, which is an important concept in stochastics.

Answer 2

Hint:

$\{ f\neq g\}= \underset{n\in \mathbb{N}}{\bigcup} \{ f-g \geq \frac{1}{n} \} \cup \underset{n\in \mathbb{N}}{\bigcup} \{ g-f \geq \frac{1}{n} \} $

Answer 3

It is enough to prove the following proposition:

If $h:(X,\mathcal F, \mu)\to \Bbb R^+_0$ is an $\mathcal F$-measurable function, then $$h\equiv 0 ~\mu -a.e.\iff \forall A\in \mathcal F:\int_A hd\mu =0.$$

We prove as follows;

If $h\equiv 0 ~\mu -a.e.$ and $A\in \mathcal F$, then $\int_A hd\mu =\int_A 0d\mu=0$. Conversely suppose $\forall A\in \mathcal F:\int_A hd\mu =0$. For $n\in \Bbb N$ we consider the sets $\{h>\frac 1n\}$ and we get $0=\int_{\{h>\frac 1n\}}hd\mu \geq \int_{\{h>\frac 1n\}}h1_{\{h>\frac 1n\}}d\mu \geq \int_{\{h>\frac 1n\}}\frac 1n d\mu =\frac 1n \mu (\{h>\frac 1n\})\implies \mu (\{h>\frac 1n\})=0.$From $\sigma$-subadditivity of $\mu$ we have $\{h>0\}=\bigcup_n \{h>\frac 1n\}\implies \mu (\{h>0\})=\mu (\bigcup_n\{h>\frac 1n\})\leq \sum_n \mu (\{h>\frac 1n\})=0$, which means that $h\equiv 0~\mu-$a.e.