Real Analysis, Folland Proposition 2.23 Integration of Complex Functions Background Information:
2.20 Proposition - If $f\in L^{+}$ and $\int f < \infty$, then $\{x:f(x) = \infty\}$ is a null set and $\{x:f(x) > 0\}$ is $\sigma$-finite.
Question:

Proposition 2.23:
a.) If $f\in L^1$, then $\{x:f(x)\neq 0\}$ is $\sigma$-finite.
b.) If $f,g\in L^1$, then $\int_{E} f = \int_{E} g$ for all $E\in M$ if and only if $\int |f - g| = 0$ if and only if $f = g$ a.e.

Attempted proof a.) Note Folland states this follows from Proposition 2.20 I will attempt to use this in part of my proof. Let $f\in L^1$ and $E = \{x:f(x)\neq 0\}$. Since $f\in L^1$ we know that $\int f < \infty$. We have from Proposition 2.20 that the set $\{x:f(x) > 0\}$ is $\sigma$-finite. Now since
$$\{x: f(x) < 0\} = \bigcup_{n\in\mathbb{N}} \{x:f(x) < 1/n\}$$
For each $n$ set $\phi_n = \frac{1}{n}\chi_{\{x:f(x)<1/n\}}$ such that $0\leq |\phi_n|\leq |f|$ Then we have
$$\int |f| \geq \int |\phi_n| = \int |\frac{1}{n}\chi_{\{x:f(x)<1/n\}}| = \frac{1}{n}\int |\chi_{\{x:f(x)<1/n\}}|$$
Thus $$n\int |f| \geq \int |\chi_{\{x:f(x)<1/n\}}|$$
Before I continue I am wondering since we already have proposition 2.20 that $\{x:f(x) > 0\}$ is $\sigma$-finite what I am trying to show above is that $\{x:f(x) < 0\}$ is $\sigma$-finite and obviously if we have $f(x) = 0$ then the quantity would equal zero thus can we conclude that $\{x:f(x)\neq 0\}$ is $\sigma$-finite.
Attempted proof b.) I need a little bit of help with this part here. Any suggestions is greatly appreciated.
 A: 
Proposition 2.23:
a.) If $f\in L^1$, then $\{x:f(x)\neq 0\}$ is $\sigma$-finite.
b.) If $f,g\in L^1$, then $\int_{E} f = \int_{E} g$ for all $E\in M$ if and only if $\int |f - g| = 0$ if and only if $f = g$ a.e.

Proof a.) Folland simply says that item a.) follows from Proposition 2.20. Here is what he probably means:
Since $f\in L^1$ we know that $\int |f| < \infty$. So, by Proposition 2.20 we have  that the set $\{x:|f(x)| > 0\}$ is $\sigma$-finite. Since $\{x:f(x)\neq 0\} = \{x:|f(x)| > 0\}$, we have that $\{x:f(x)\neq 0\}$ is $\sigma$-finite.
Proof b.) In item b.) must prove that, if $f,g\in L^1$, then the three statements below are equivalent:

*

*$\int_{E} f = \int_{E} g$ for all $E\in M$


*$\int |f - g| = 0$


*$f = g$ a.e.
Follands says that the equivalence of 2 and 3 follows from Proposition 2.16. In fact, note that $|f-g| \in L^+$, so by Proposition 2.16, we have $\int|f-g| =0$ iff $|f-g|=0$ a.e.. Since   $|f-g|=0$ a.e. iff $f=g$ a.e., we have that  $\int|f-g| =0$  iff  $f=g$ a.e.
Now let us complete the proof of item b.).
(2 $\Rightarrow$ 1) Suppose $\int |f-g| =0$. By proposition 2.22 we have, for any $E \in \mathcal{M}$,
$$\left | \int_E f - \int_E g \right | \leq \left | \int \chi_E (f -  g) \right|\leq \int |  \chi_E (f -  g) | \leq  \int | f -  g |$$
So $\int_E f = \int_E g$.
(1 $\Rightarrow$ 2) Here I will prove in a way slightly different from the proof in Folland's book.
Suppose that  $\int_{E} f = \int_{E} g$ for all $E\in M$. Since $f, g \in L^1$, we have that, for all $E\in M$,  $\int_{E} f <\infty$ and
$\int_{E} g <\infty$. So we have that, for all $E\in M$,
$$\int_{E} (f -g) =\int_{E} f - \int_{E} g =0$$
Let $A =\{x: (f-g)(x) \geq 0\}$. We have
$$\int |f - g| = \int ( \chi_A(f-g) - \chi_{A^c} (f-g) ) = \int_A (f-g) - \int_{A^c} (f-g) =0 - 0=0$$
A: For part $(a)$, if $f\in L^1$, you can apply Proposition $2.20$ to $\lvert f\rvert$ to deduce that the set $\{x : \lvert f(x)\rvert > 0\}$ is $\sigma$-finite. But this set is the same as $\{x : f(x) \neq 0\}$.
For part $(b)$, suppose $\int_E f\, dm = \int_E g\, dm$ for all $E\in M$. Let $E_n := \{x\in X : f(x) - g(x) > 1/n\}$ and $F_n :=\{x\in X : f(x) - g(x) < -1/n\}$, for $n = 1,2,3,\ldots$. Then $$\int_{E_n} \lvert f - g\rvert = \int_{E_n} (f - g)\, dm = \int_{E_n} f\, dm - \int_{E_n} g\, dm = 0$$ and $\int_{F_n} \lvert f - g\rvert\, dm = \int_{E_n} (g - f)\, dm = \int_{E_n} g\, dm - \int_{E_n} f\, dm = 0$. Thus $\int_{G_n} \lvert f - g\rvert \, dm = 0$, where $G_n :=\{x \in X : \lvert f(x) - g(x)\rvert < 1/n\}$. Let $G = \{x\in X : f(x) - g(x) \neq 0\}$. As $f = g$ on $G^c$, $\int_{G^c} \lvert f - g\rvert\, dm = 0$. Since $G$ is the union of the sets $G_n$ and $\int_{G_n} \lvert f - g\rvert\, dm = 0$, then $\int_G \lvert f - g\rvert \, dm = 0$. Thus $\int \lvert f - g\rvert = 0$.
If $\int \lvert f - g\rvert\, dm = 0$, then $m(G_n) \le n\int \lvert f - g\rvert\, dm = 0$, so $G_n$ is a null set. Since $G$ is the countable union of null sets $G_n$, then $G$ is a null set. Therefore, $f = g$ a.e..
Finally, if $f = g$ a.e., then given $E\in M$, $m(E\cap G) = 0$ and $f = g$ on $E\cap G^c$; therefore, $\int_{E\cap G} (f - g)\, dm = 0$ and $\int_{E\cap G^c} (f - g)\, dm = 0$. Consequently, $$\int_E (f - g)\, dm = \int_{E\cap G} (f - g)\, dm + \int_{E\cap G^c} (f - g)\, dm = 0$$ that is, $\int_E f\, dm = \int_E g\, dm$. This proves the equivalence in part (b).
