Suppose that $f_{1}$ and $f_{2}$ are in $L(X,\Omega, \mu)$ and let $\lambda_{1}$ and $\lambda_{2}$ be their indefinite integrals. Show that $\lambda_{1}(E)=\lambda_{2}(E)$ for all $E\in \Omega$ if and only if $f_{1}(x)=f_{2}(x)$ for almost all $x\in \Omega$.
My attempts: I've proved the first implication, this is my idea:
$f_{1}$ and $f_{2}$ are in $L(X, \Omega, \mu)$ then we know that: $$\displaystyle\int f_{1}d\mu=\displaystyle\int f_{1}^{+}d\mu-\displaystyle\int f_{1}^{-}d\mu$$ and $$\displaystyle\int f_{2}d\mu=\displaystyle\int f_{2}^{+}d\mu-\displaystyle\int f_{2}^{-}d\mu$$ And if $\lambda_{1}$ and $\lambda_{2}$ are the indefinite integrals of $f_{1}$ and $f_{2}$ then we know that: $$\lambda_{1}(E)=\displaystyle\int\limits_{E}f_{1}d\mu=\sum\limits_{n=1}^{\infty}\int\limits_{E}f_{1}d\mu$$ And the same with $\lambda_{2}$ and $f_{2}$, then if we asume that $\lambda_{1}(E)=\lambda_{2}(E)$ using the $\lambda$ definition we can show that $f_{1}(x)=f_{2}(x)$. But I don't know how to prove the counterimplication, I mean, when we suppose that $f_{1}(x)=f_{2}(x)$. If somebody could help me, I'll be so grateful.
If $X$ is sigma finite and $f_1,f_2$ are real-valued, you can argue as follows:
let $E=\left \{ f_1\ge f_2 \right \}.$ Then, $\int_Ef_1-f_2=0\Rightarrow f_1-f_2=0$ ae $E$. Similarly, if $F=\left \{ f_1\le f_2 \right \},\ $ we have $f_2-f_1=0$ ae $F$ and so conclude that $f_1=f_2$ ae $E\cup F=X.$
For the complex case, consider real and imaginary parts.
$*$ We used the fact that If $f$ is a nonnegative Lebesgue integrable function defined on a Lebesgue measurable set $E$ and $\int_Ef=0$ then $f=0$ almost everywhere on $E.$ The proof goes as follows:
First, suppose $\lambda (E)<\infty.$Then, define $E_n=\left \{ f>1/n \right \}.$ Then, $\lambda (E_n)/n=\int_{E_n}1/n\le \int_{E_n}f=0\Rightarrow \lambda (E_n)=0.$ To finish, note that the $E_n$ increase to $\left \{ f>0 \right \}$. I'll leave the The sigma finite case to you.