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:


$S := \{x \in \Omega: f(x) \neq 0\}$. 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.


$\phi_k \uparrow X_s$, 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.

  • $\begingroup$ Why not just use Markov's inequality? $\endgroup$ – BCLC May 3 '18 at 14:35

Another way to write $\phi_k$ is $$ \phi_k(x) = \begin{cases} 0 & \text{if $f(x) = 0$}\\ k|f(x)| & \text{if $|f(x)| \leq \frac{1}{k}$} \\ 1 & \text{if $|f(x)| \geq \frac{1}{k}$} \end{cases} $$ So if $f(x) = 0$, $\phi_k(x) = 0 = \chi_S(x)$ for all $k$. If $f(x) \neq 0$, there exists $K$ such that $|f(x)| > \frac{1}{K}$. Then for all $k \geq K$, $\phi_k(x) = 1 = \chi_S(x)$.

  • $\begingroup$ Why not just use Markov's inequality? $\endgroup$ – BCLC May 3 '18 at 14:36
  • $\begingroup$ @BCLC If I understand correctly, the OP was trying to understand a specific proof of this fact, not prove it on their own. $\endgroup$ – Matthew Leingang May 5 '18 at 12:05

Let $x \in \Omega$.

Case 1: $x \notin S$. Then $f(x)=0=X_S(x)$, hence: $\phi_k(x)=0=X_S(x)$ for all $k$.

Case 2: $x \in S$. Then $|f(x)|>0$. Hence there is $k_0$ such that $k|f(x)| \ge 1=X_S(x)$ for all $k>k_0$. Then we have

$\phi_k(x)=X_S(x)$ for all $k>k_0$.

It is your turn to show that $\phi_1(x) \le \phi_2(x) \le ....\le \phi_{k_0}(x)$

  • $\begingroup$ Why not just use Markov's inequality? $\endgroup$ – BCLC May 3 '18 at 14:35

Reverse of standard machine:

$$\int_{\Omega} |f| d\lambda = 0$$

$$\to \int_{\Omega} f^+ d\lambda + \int_{\Omega} f^- d\lambda = 0$$

$$\to \int_{\Omega} f^+ d\lambda = \int_{\Omega} f^- d\lambda = 0$$

$$\to \sup_{h^+ \le f^+} \int_{\Omega} h^+ d\lambda = \sup_{h^- \le f^-} \int_{\Omega} h^- d\lambda = 0$$

$$\to \int_{\Omega} h^+ d\lambda = \int_{\Omega} h^- d\lambda = 0$$

$$\to \int_{\Omega} \sum_{i=1}^n a_i 1_{A_i} d\lambda = \int_{\Omega} \sum_{j=1}^n b_j 1_{B_j} d\lambda = 0$$

$$\to \sum_{i=1}^n a_i \lambda(A_i) = \sum_{j=1}^n b_j \lambda(B_j) = 0$$

$$\to a_i \lambda(A_i) = 0, b_j \lambda(B_j) = 0$$

$$\to a_i = 0 \ or \ \lambda(A_i) = 0, b_j \ or \ \lambda(B_j) = 0$$

$$\to \lambda(\{x \in \Omega: h^+ \neq 0\})=\lambda(\{x \in \Omega: h^- \neq 0\}) = 0$$

$$\to \lambda(\{x \in \Omega: f^+ \neq 0\})=\lambda(\{x \in \Omega: f^- \neq 0\}) = 0$$

$$\to \lambda(\{x \in \Omega: f \neq 0\}) = 0$$



By Markov's inequality, $\forall \varepsilon > 0$,

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

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


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