almost every where property I have question ,let $f$ is measurable and integrable i.e $f\in(\Omega,\mathbf{A}, \mathbb{R})$ and for all $A\in \mathbf{A}$  $\int_Af(\omega)d\mu=0$ show that $f=0$  almost every where.
answer: I wil solve it from couter example sps that f is not 0 ie $f(\omega)>0$so I define the set ${\omega \in \Omega :f(\omega)>0}$ and I know that the set ${ \omega \in\Omega :f(\omega)>\epsilon}$ is a subset of $f(\omega)>0$ and from hypothesis of question I know that $\int f(\omega)>0=0$ so the $\int f(\omega)<\epsilon$ is also 0 how I can cuncluderen that $f(\omega)=0$ almost every where
 A: Why don't you apply the hypothesis to 
$$
A = \{\omega:f(\omega) <0\}
\\
B = \{\omega:f(\omega) >0\}
$$?
A: Let $\epsilon>0,A_{\epsilon}=f^{-1}([\epsilon,\infty])$, and note the $\epsilon\chi_{A_\epsilon}\leq f$ is a simple function.
$$\epsilon\mu(A_{\epsilon})=\int_{A_{\epsilon}}\epsilon\chi_{A_{\epsilon}}d\mu\leq\int_{A_{\epsilon}}fd\mu=0$$ gives $\mu(A_{\epsilon})=0$ for every $\epsilon>0$. Furthermore, $f^{-1}((0,\infty])=\bigcup_{n=1}^{\infty}A_{\frac{1}{n}}$. So we have $$0\leq\mu(f^{-1}((0,\infty]))=\mu\left(\bigcup_{n\geq 1}A_{\frac{1}{n}}\right)\leq\sum_{n=1}^{\infty}\mu(A_{\frac{1}{n}})=0.$$ Thus $f\leq 0$ almost everywhere (on a set $E^{+}$ such that $\mu(\mathbb{R}\setminus E^{+})=0$).
Replacing $f$ with $-f$ in the preceding argument yields $-f\leq0$ (aka $f\geq0$) almost everywhere (this time on a set $E^{-}$). Thus we have $0\leq f\leq 0$ on all but a set of measure zero -- namely $f=0$ on $E^{+}\cap E^{-}$, and $$\mu(\mathbb{R}\setminus(E^{+}\cap E^{-}))=\mu((\mathbb{R}\setminus E^{+})\cup(\mathbb{R}\setminus E^{-}))\leq \mu(\mathbb{R}\setminus E^{+})+\mu(\mathbb{R}\setminus E^{-})=0$$.
