$\int_{\Omega} |f| d\lambda = 0 \Rightarrow \{x \in \Omega: f(x) \neq 0\}$ is a null set 
Let $(\Omega, \mathscr A, \lambda)$ be a measure space.
For an $\mathscr A$-measurable numerical function $f: \Omega \rightarrow \Bbb R$, it holds that
$\int_{\Omega} |f| d\lambda = 0 \Rightarrow \{x \in \Omega: f(x) \neq 0\}$ is a null set.

The argument works the following way:
Define
$S := \{x \in \Omega: f(x) \neq 0\}$. Since $f$ is measurable, it follows that $S \in \mathscr A$. For every natural number $k \ge 1$, we define
$\phi_k := \inf\{k|f|, X_s\}$ with $X_s$ being the Indicator function.
Hence,
$\phi_k \uparrow X_s$, and since $\phi_k \le k|f|$, we get $\int \phi_k d\lambda = 0$.
While I understand the very last step here, I don't see why $\phi_k \uparrow X_s$. $X_s = 0$ for every $x \notin S$, and since $S$ is defined the way it is, it's only possible that $|f| = 0$, and therefore, $k|f| = 0$ for every $k$. So how does $\phi_k$ converge against $X_s$ here?

Reverse of standard machine:
$$\int_{\Omega} |f| d\lambda = 0$$
$$\to \int_{\Omega} f^+ d\lambda + \int_{\Omega} f^- d\lambda = 0$$
$$\to \int_{\Omega} f^+ d\lambda = \int_{\Omega} f^- d\lambda = 0$$
$$\to \sup_{h^+ \le f^+} \int_{\Omega} h^+ d\lambda = \sup_{h^- \le f^-} \int_{\Omega} h^- d\lambda = 0$$
$$\to \int_{\Omega} h^+ d\lambda = \int_{\Omega} h^- d\lambda = 0$$
$$\to \int_{\Omega} \sum_{i=1}^n a_i 1_{A_i} d\lambda = \int_{\Omega} \sum_{j=1}^n b_j 1_{B_j} d\lambda = 0$$
$$\to \sum_{i=1}^n a_i \lambda(A_i) = \sum_{j=1}^n b_j \lambda(B_j) = 0$$
$$\to a_i \lambda(A_i) = 0, b_j \lambda(B_j) = 0$$
$$\to a_i = 0 \ or \ \lambda(A_i) = 0, b_j \ or \ \lambda(B_j) = 0$$
$$\to \lambda(\{x \in \Omega: h^+ \neq 0\})=\lambda(\{x \in \Omega: h^- \neq 0\}) = 0$$
$$\to \lambda(\{x \in \Omega: f^+ \neq 0\})=\lambda(\{x \in \Omega: f^- \neq 0\}) = 0$$
$$\to \lambda(\{x \in \Omega: f \neq 0\}) = 0$$
QED
