I know the proof of this question:
Let $f$ be a nonnegative bounded measurable function on a set of finite measure $E$. Assume $\int_{E} f = 0.$ Show that $f = 0$ a.e. on $E.$
Here it is.
Let $f$ be a bounded measurable function on a set of finite measure $E.$Then by $\textbf{Theorem 4 on page 74}$, $f$ is integrable over $E.$\
Also, By $\textbf{page 54 definition of a.e.}$, we know that a property is said to hold almost everywhere (abbreviated a.e.) on a measurable set $E$ provided it holds on $E \setminus E_{0}$, where $E_{0}$ is a subset of $E$ for which $m(E_{0}) = 0.$\
Now, since $f$ is nonnegative, so $f$ can not take negative values and we have only 2 sets:
Define $$A :=\{ x\in E: f(x)=0 \}$$ and $$B:=\{x\in E: f(x)>0\}.$$ Then $$\int_E f = \int_A f + \int_B f = \int_B f = 0.$$\ because $f(x) = 0$ on $A$ and using $\textbf{page 76 corollary 6,}$ as $E = A \cup B$ and the last equality because we have an assumption of $ \int_{E}f = 0.$\
Since $f$ is positive on $B,$ in order for $\int_B f=0,$ one must have $m(B)=0,$ that is, Lebesgue measure of $B$ must be zero.\
\textbf{Now to show that $m(B)=0.$}\
Since $B$ is measurable, we can write it as a countable union of disjoint measurable sets. For example $B = \bigcup_{n=1}^{\infty}B_n$ where $$B_n = \{x \in E:f(x)\ge 1/n\}.$$
Now it is enough to prove that the measure of each of the sets in the union is zero. Let there be a natural number $N$ such that $$m( \{x \in E:f(x)\ge 1/N\}) = \lambda (\neq 0).$$
Now, $Let E_{1} = \{x \in E :f(x)\ge 1/N \}, $ Let $E_{2} = \{x \in E :f(x)\leq 1/N \}$, thus $E_{1}$ and $E_{2}$ are measurable disjoint sets and $E = E_{1} \cup E_{2}.$ therefore $$ \int_{E}f = \int_{E_{1}}f + \int_{E_{2}}f.$$\
But $\int_{E_{1}}f > (1/N) m({E_{1})$, because $f(x)\ge 1/N$ on $E_{1}.$ But $m({E_{1}) = \lambda ,$ therefore $\int_{E_{1}}f = (\lambda / N) > 0.$ Consequently $\int_{E} f > 0,$ which contradicts the hypothesis that $\int_{E} f = 0.$\
Hence, $m(B):=\{x\in E: f(x)>0\} = 0$ therefore $f = 0$ a.e on $E$.
But I do not know how to prove this question:
Let $f$ be integrable over $X$ with respect to $\mu.$ Show that $\int_{E}f d\mu = 0$ for every measurable subset $E$ of $X$ if and only if $f = 0$ a.e. on $X.$
I got a hint here How are these two proofs should differ? but still I do not know how to apply it.
My confusion is:
$\int_{E}fd\mu $ could be zero even though one of the sets in the union mentioned in the hint have positive measure and another set in the same union have the same value of measure but with negative sign ...... or the negative sign measure is not allowed?
I am confused about how to implement the hint, could anyone help me in doing so please?
If $f\neq 0$ on $E$ with $\mu E>0$, let $E_+=\{x\in E:f(x)>0\},\, E_- =\{x\in E: f(x)<0\}$. Then $E$ is the disjoint union of $E_+$ and $E_-$, so one of which has positive measure.