Let $f:\mathbb{R}\rightarrow[0,\infty]$ be measurable and assume that $\int_{\mathbb{R}}f(x)dx=0$. Prove that $f(x)=0$ for a.e. $x\in\mathbb{R}$.
I know the proof is simple if the function $f$ is defined to be continuous because then, we can show that the sign is preserved. But in this case, I'm not to assume that the given function is continuous. How can I go about proving this?
Let $E_n=\{x~:~f(x)\geq \frac{1}{n}\}$. Since $f$ is non-negative measurable function and $\int f\,dx=0$,
$$\int f\,dx\geq \int \frac{1}{n}~ \chi_{E_n}\,dx= \frac{1}{n}~m(E_n).$$ $$\implies m(E_n)=0$$ $$\{x~|~f(x)>0\}=\cup_{n=1}^\infty E_n.$$ We know that if $\{E_i\}$ is a sequence of measurable sets such that $E_1\subseteq E_2\subseteq E_3\subseteq \cdots$, then $m(\{x~|~f(x)>0\})=m(\cup_{n=1}^\infty E_n)=m(lim E_n)=lim~ m(E_n)=0$. Hence, $f=0$ a.e. .