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!


2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$
    – meth-
    Oct 20, 2022 at 11:39
  • 1
    $\begingroup$ This may help: math.stackexchange.com/questions/2905407/… $\endgroup$ 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 $\int|f|\,d\mu=\int_{\{|f|>0\}}|f|\,\mu=0$.


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