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Let $X$ be a locally compact Hausdorff space, and let $\mu$ be a regular measure on $X$. Suppose that $g : X \to \Bbb C$ belongs to $L^{\infty}(X)$.

My question is :

Is it sufficient to assume that $\int_E g = 0$ for all Borel sets $E$ of finite measure (that is $\int_X fg = 0$ for all $f \in L^1(X)$) to conclude that $g=0$ a.e. ?

I know from questions like this one that this is true if we assume $\int_E g = 0$ for all Borel sets $E$.

Given a Borel set $E$ of infinite measure, maybe I can write it as a countable union of Borel sets of finite measure to use my hypothesis. (Or the answer to my question could be "no", but this would be annoying for me...)

Thank you very much for your help!

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  • $\begingroup$ If you multiply by indicator functions, you can compare the two cases. $\endgroup$ – Michael Burr May 1 '16 at 18:25
  • $\begingroup$ @MichaelBurr : so you mean that for any $x \in X$, I take a Borel set $A$ of finite measure that contains $x$, and then $g \chi_A$ has zero integral over every Borel set, so that $g$ is zero a.e. on $A$? How to conclude that $g$ is zero a.e. on $X$? $\endgroup$ – Alphonse May 1 '16 at 18:41
  • $\begingroup$ By changing $A$, you can get that it is zero a.e. on $X$. $\endgroup$ – Michael Burr May 1 '16 at 19:32
  • $\begingroup$ @MichaelBurr : yes, but isn't there a problem if I need uncountably many such $A$'s of finite measure? An uncountable union of sets of measure zero may not be zero (even not measurable). $\endgroup$ – Alphonse May 1 '16 at 19:53
  • $\begingroup$ Agreed, this will work if $X$ can be written as a union of countably many finite measure subsets. If $X$ cannot be written in this way, I would expect that the conditions are not equivalent (just as you say, by setting $g$ to be nonzero on uncountablly many measure zero sets). $\endgroup$ – Michael Burr May 1 '16 at 19:58
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If $X$ is not $\sigma$-finite, the answer is NO. Here is a simple example.

Let $X=[0,1]$ with the usual topology. Let $\mu$ be defined on the Borel $\sigma$-algebra, by $\mu(\emptyset)=0$ and $\mu(E)=+\infty$ if $E\neq \emptyset$.

It is easy to see that$\mu$ is a regular measure. Let $g$ be the constant function, $g=1$.

We have $\int_E g = 0$ for all Borel sets $E$ of finite measure (because the only Borel set of finite measure is the empty set!) and, it is clear that we can not conclude $g=0$ a.e.

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If $\mu$ is a Radon measure (which is not an inappropriate hypothesis for a measure on a locally compact Hausdorff space) then the assertion is true. To see this fix $\epsilon>0$ and define $B:=\{x\in X: g(x)>\epsilon\}$. Because $\mu$ is Radon there is a sequence $\{K_n\}$ of compact subsets of $B$ with $\sup_n\mu(K_n)=\mu(B)$. Because $K_n$ is compact, $\mu(K_n)<\infty$. Therefore $$ 0=\int_{K_n}g\,d\mu\ge \epsilon\mu(K_n). $$ It follows that $\mu(B)=0$. Sending $\epsilon$ down to $0$ along a sequence, we obtain $\mu(\{x:g(x)>0\})=0$. Likewise, $\mu(\{x:g(x)<0\})=0$. That is, $g=0$, $\mu$-a.e.

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  • $\begingroup$ Thank you very much! This is really helpful. Actually I had in mind a Radon measure instead of a regular measure… I'm just wondering : why are you saying "which is not an inappropriate hypothesis for a measure on a locally compact Hausdorff space"? For me this is appropriate (e.g. Haar measure on a locally compact group). $\endgroup$ – Alphonse May 2 '16 at 6:43

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