$\textbf{Problem}$ Let $L=L(X,\textbf{X},\mu)$ denote the set of all integrable functions. Suppose that $f$ belongs to $L$ and that its indefinite integral is $$\lambda(E)=\int_{E} f d\mu, E\in \textbf{X}.$$ Show that $\lambda(E)\geq 0$ for all $E\in\textbf{X}$ if and only if $f(x)\geq 0$ for almost all $x \in X$.
$\textbf{Solution}$ Assume $f(x)\geq 0$ for almost all $x\in X$. Pick any set $E\in\textbf{X}$. There are two cases to consider. If $f(x)\geq 0$ for all $x\in E$, then $f^{+}\chi_{E} \geq f^{-}\chi_{E}$. Hence, $\int_{E} f^{+} d\mu \geq \int_{E} f^{-} d\mu$; therefore, $$\lambda(E)=\int_{E} f d\mu =\int_{E} f^{+} d\mu - \int_{E} f^{-} d\mu \geq 0.$$ If there exists an $x\in E$ such that $f(x)<0$ then because $f(x)\geq 0$ for almost all $x\in X$, $E$ must have measure zero. Since, in this case, both functions $f^{+}\chi_{E}$ and $f^{-}\chi_{E}$ would be $0$ almost everywhere, their integrals are zero. Thus, $\lambda(E)=\int_{E} f d\mu = 0$.
I am having trouble proving the other direction. I want to pick any set $E$ in our $\sigma$-algebra X and show that (1) if $E$ does not have measure zero, then we must have $f(x)\geq 0$, and (2) if $E$ does have measure zero, then we cannot determine anything about $f(x)$.
I think I got (2). I am just unsure about (1). A hint would be great! Thank you!
For the first part, what is implicitly used is that if $\mu(N)=0$ then for each integrable function $f$, $\int_Nf\mathrm d\mu=0$.
For the converse, as Did suggested, fix a positive $n$. With the choice $E:=\{x,f(x)\leqslant -1/n\}$, we get $$0\leqslant\lambda(E)=\int_Ef(x)\mathrm d\mu\leqslant -\frac 1n\mu(E).$$ As $\mu$ is a positive measure, $\mu(E)=0$. Now, notice that $\{f<0\}=\bigcup_{n>0}\{x,f(x)\leqslant-\frac 1n\}$, a countable union of sets of measure $0$.