# Showing that if $f$ vanishes locally almost everywhere then integral of $|f|$ is 0

I am studying measure theory from Cohn's Measure theory textbook. If $$(X, \scr A, \mu)$$ is a measure space then a subset $$N$$ of $$X$$ is said to be locally null if for every $$A \in \scr A$$ with $$\mu (A) < + \infty$$, we have that $$A \cap N$$ is null set. Also, subset $$B$$ of $$X$$ is said to be null set if there is a set $$A \in \scr A$$ such that $$B \subset A$$ and $$\mu (A) = 0$$.

Now, here's what I am trying to prove: if $$f: X \to \mathbb C$$ and $$f=0$$ locally almost everywhere, that is, $$\{ x\in X : |f(x)| >0 \}$$ is locally null then $$\int |f| \,d\mu = 0$$. This claim would be true if $$f=0$$ almost everywhere, that is, the $$\{ x\in X : |f(x)| >0 \}$$ is null.

This holds true if $$X$$ is $$\sigma$$-finite. Because $$\sigma$$-finiteness would imply that every locally null set is null and we would be done.

The claim remains to be proved when $$X$$ is not $$\sigma$$-finite. I tried my best to prove it but could not reach anywhere. Hints to prove or disprove it will be appreciated!

You cannot prove this. $$\int |f|d\mu=0$$ if and only if $$f=0$$ a.e. Take a locally null set $$A$$ which is not a null set and take $$f=\chi_A$$. This is a counter-example to your statement.

• Here $A$ is required to be in the sigma-algebra for otherwise $\chi _ A$ may not be measurable. So is there a measure space where the locally null set which is not a null set a member of the sigma-algebra? Oct 20, 2022 at 11:39
• This may help: math.stackexchange.com/questions/2905407/… Oct 20, 2022 at 11:41

The statement in the title of the OP's posting is not quite correct. Here is a a counter example: consider the set $$[0,1]$$ equipped with the Borel $$\sigma$$-algebra, and define the measure $$\mu$$ as $$\mu=\infty\cdot\delta_0+\mathbb{1}_{(0,1]}\cdot\lambda$$ where $$\delta_0$$ is the measure concentrated at $$0$$ and $$\lambda$$ is Lebesgue measure. Notice that $$\mu(\{0\})=\infty$$ and $$\mu((0,1])=1$$. Let $$f(x)=\mathbb{1}_{\{0\}}$$. This function is locally null for $$\{|f|>0\}=\{0\}$$, and for any set $$A$$ with finite $$\mu$$ measure $$\{|f|>0\}\cap A=\emptyset$$ and so $$\mu\big(\{|f|>0\}\cap A\big)=0$$. However $$\int_{[0,1]}f\,d\mu=\infty$$.

The issue with this measure is that it there is an atom $$\{0\}$$ of infinite mass. Measures without this pathology are called semifinite measures. To be more precise

A measure $$\mu$$ on a measurable space $$(X,\mathscr{B})$$ is semi finite of for any $$B\in\mathscr{B}$$, if $$\mu(B)>0$$, then there is $$A\in \mathscr{B}$$ with $$A\subset B$$ and $$0<\mu(A)<\infty$$.

Any $$\sigma$$-finte measure is semidefinite. We have the following result

If $$(X,\mathscr{B},\mu)$$ is a semifinte measure space and $$f$$ is a locally null measurable function, then $$\int|f|=0$$.

Here is a sketch o a proof: For any set $$E\in\mathscr{B}$$ define $$\mu_0(E)=\sup\{\mu(F): F\in\mathscr{B},\, F\subset E, \,\mu(B)<\infty\}$$ It is not difficult to check that $$\mu_0(E)=\mu(E)$$ whenever $$\mu$$ is semifinite (see Exercises 14, 16 on pages 27 of Folland. G., Real Analysis, 2nd edition, Wiley & Sons). Thus, if $$f$$ is locally null, then $$\mu(|f|>0)=\mu_0(|f|>0)=0$$; hence $$\int|f|\,d\mu=\int_{\{|f|>0\}}|f|\,\mu=0$$.