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This question appeared on a Measure Theory exam a couple days ago:

Let $(X,\mathscr M,\mu)$ be a measure space. Suppose $\mathbf{f:X\to[0,\infty)}$ is positive and measurable. Define a measurable set

$$\mathbf{E:=\{x\in X:f(x)>0\}}$$

Prove

$$\mathbf{\lim_{n\to\infty}\int f(x)^{1/n}d\mu(x)=\mu(E)}$$

Can someone explain to me why the following is not a counterexample:

Let $X=\mathbb R$, $\mathscr M=\mathscr B_{\mathbb R}$, $\mu$ be the usual Borel measure such that $\mu((a,b))=b-a$, and

$$f(x)=\left\{ \begin{array}{lr} e^{1/x}& \text{if } x\in (0,1)\\ 0& \text{if } x\notin (0,1) \end{array} \right.\\$$

$f$ is clearly positive, and it is measurable because it can be written as $e^{1/x}\chi_{(0,1)}(x)$, $e^{1/x}$ is continuous everywhere where it is relevant, and $\chi_{(0,1)}$ is the characteristic function of a Borel-measurable set, so $f$ is the product of measurable functions. $E=(0,1)$ and $\mu(E)=1$. However, for all $n>0$,

$$\int f(x)^{1/n}d\mu(x)=\int_{(0,1)}\left(e^{1/x}\right)^{1/n}d\mu(x)=\int_{(0,1)}e^{1/nx}d\mu(x)=\infty$$

$$lim_{n\to\infty}\int f(x)^{1/n}d\mu(x)=\infty\ne 1=\mu(E)$$

It seems to me you need some kind of limit on $f$ that allows you to prove that, for some $n>0$, $\int f(x)^{1/n} d\mu(x)<\infty$. Measurability is not enough to assure that.

Thanks for any help!

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    $\begingroup$ I haven't looked carefully at your example, but yes, the statement is false without some extra condition - it's easy to give examples where all those integrals are infinite but $E$ has finite measure, as you claim to have done. $\endgroup$ Apr 21, 2016 at 13:38

1 Answer 1

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FWIW: The professor agrees that the original question was in error: one can't get the result without some other sort of limit on $f$.

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