$\int_{\Omega} |f| d\lambda = 0 \Rightarrow \{x \in \Omega: f(x) \neq 0\}$ 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?
From construction it is clear that $\phi_k \leq \chi_S$ for all $k$. To see that $\phi_k$ converges pointwise to $\chi_S$ consider first if $x\in S$. Then $f(x)\neq 0$ and so $k |f(x)|\geq 1$ for sufficiently large $k$. Thus, eventually $k|f(x)|\geq \chi_S(x)$ and $\phi_k(x) = \chi_S(x)$. On the other hand, if $x\not\in S$ then $f(x)=0$ and $\chi_S(x)=0$, so it is clear that $\phi_k(x)  = \chi_S(x)$ for all $k$.

Together, these two cases imply that $\phi_k$ converges to $\chi_S$ pointwise.
By Markov's inequality, $\forall \varepsilon &gt; 0$,

$$P(f \ne 0) = P(|f| \ge \varepsilon) \le \frac{E[|f|]}{\varepsilon} = 0$$

I know yours is a measure but not probability space but the same holds true for measure space.