I would like to prove that for a given measure space $(E,\mathcal{E},\mu)$ and a measurable function with values in the extended positive real line equipped with the sigma algebra generated by the borel sets we have :
$$ \int_{E}fd\mu = 0 \implies f = 0\quad a.e $$
We know that there exists a increasing sequence of simple functions converging to $f$ that is :
$$ \forall x\in E,\forall\varepsilon>0, \exists N\in\mathbb{N} : n>N\implies \lvert f_n(x) - f(x)\rvert<\varepsilon $$
We have that $\forall n\in\mathbb{N} : 0\leq f_n \leq f_{n+1}$ which implies
$$ \int_{E}f_nd\mu\leq\int_{E}f_{n+1}d\mu $$
By the monotone convergence theorem, we know that $(\int_{E}f_{n}d\mu)_{n}$ converges to $\int_{E}fd\mu$ but since the sequence of integrals is increasing, this means that
$$ \forall n\in\mathbb{N} : \int_{E} f_nd\mu = 0\implies \sum_{i=1}^{m}\alpha_i^{n}\mu(A_i^{n}) = 0 $$
where $A_i^{n} = \{x\in E : f(x) = \alpha_i^{n}\}$. Since the simples functions are positive, this means that if $\alpha_i^{n}>0$ we have $\mu(A_i^{n}) =0$ which means that
$$ \forall n\in\mathbb{N} : f_n = 0\quad a.e $$
Now I would like to pursue the proof, but I realized that I am not convinced by the argument I wanted to use : conclude that since $f_n$ is a sequence of null functions, it converges to the null function, that is $f=0$ a.e.
So I would like to have your help on this please, in what direction can I go at this step ?
Thank you !
Thanks to geetha290krm here is the end of the proof :
If $x\in E$ is such that $f(x)>0$, then there exists $n\in\mathbb{N}$ such that $f_n(x)>0$. Thus we have the following implications
$$ x\in \{ x\in E : f(x)>0\}\implies\exists n\in\mathbb{N} : f_n(x)>0 $$
which is equivalent to
$$ \exists n\in\mathbb{N} : x\in \{x\in E : f_n(x) > 0\}\Longleftrightarrow x\in\bigcup_{n\in\mathbb{N}}\{ x\in E : f_n(x)>0\} $$
Thus
$$ \{ x\in E : f(x)>0\}\subset\bigcup_{n\in\mathbb{N}}\{ x\in E : f_n(x)>0\} $$
We see that $\{f_n>0\}\subset\{f_{n+1}>0\}$ since the sequence is increasing. It follows that
$$ \mu(\bigcup_{n\in\mathbb{N}}\{f_n>0\}) =lim_{n\to+\infty}\mu(\{f_n>0\}) = 0 $$
since we have shown that for all $n\in\mathbb{N}$ $f_n=0$ a.e. Hence we have our conclusion since $0\leq\mu(\{f>0\})\leq \mu(\bigcup_{n\in\mathbb{N}}\{f_n>0\}) = 0$