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I have this probably silly doubt about what is $\displaystyle\lim_{n \to \infty} \int_{\mathbb{R}} \dfrac{1}{n} \mathrm{d}m$.

Here I have the Lebesgue integral.

If I'm calculating correctly, each integral in the sequence is $\infty$, so I have a constant sequence of $\infty$, and a constant sequence converges to any of its elements. (?) (This may fail for infinity, but I don't know how to verify.)

Looking at other way, I can take $\dfrac{1}{n}$ out of the integral, and $\int_{\mathbb{R}^n}1\mathrm{d}m=\infty$, but $\lim \frac{1}{n} = 0$, so I have $0.\infty$, which in measure theory is $0$.

I'm really confused, any help would be appreciated.

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    The value of the integral is $\infty$ for all $n$. – copper.hat Mar 16 '21 at 20:12
  • So the limit is also $\infty$? My problem is I don't know what is the limit of a "constant" sequence of infinity –  Mar 16 '21 at 20:14
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    The Integral is $\infty$ for all $n \in \mathbb N$. Then taking the limit $n \to \infty$ doesn't change anything because the value is independent of $n$. – Lukas Mar 16 '21 at 20:16

2 Answers2

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Even if you take the $\frac1n$ out of the integral, you have $$ \lim_{n\to\infty}\int_{\Bbb R^n}\frac1ndm =\lim_{n\to\infty}\left(\frac1n\int_{\Bbb R^n}dm\right) =\lim_{n\to\infty}\frac\infty n =\lim_{n\to\infty}\infty =\infty. $$

Rewriting $\lim_{n\to\infty}\left(\frac1n\int_{\Bbb R^n}dm\right)$ as $(\lim_{n\to\infty}\frac1n)(\lim_{n\to\infty}\int_{\Bbb R^n}dm)$ is only valid if the resulting product (in this case $0\times\infty$) is not indeterminate. See: when is the limit of a product the product of the limits?

Karl
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In measure theory, there are some important theorems that justify interchange $\lim \int = \int \lim$. But the hypotheses of those theorems are important, too. Your example does not satisfy the needed hypotheses.

GEdgar
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