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Let $(\Omega, \Sigma, \mu)$ be a measure space. It is easy to prove that a uniform limit $f$ of real-valued functions $f_k \in \mathcal{L}(E)$ is also integrable and $\lim \limits_{k \to \infty} \int_{E} f_k d\mu= \int_{E} f d\mu.$

Now there is an exercise in my lecture notes that asks me to prove that this fails if $\mu(E)=\infty$. To show this I am supposed to find a sequence of functions $f_k \in \mathcal{L}([1,\infty))$ such that $f_k$ converges uniformly to $f$ and $f \notin \mathcal{L}([1,\infty))$.

The lecture notes give the following hint:

$f = \frac{1}{\sqrt{x}}$

How can I show this? I don't really know how to start. To prove that $f$ is not integrable I need to show that its integral is infinite (because $f$ is continuous, hence measurable).

Can anybody help me out please?

Thanks a lot!

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The function $f(x) = 1/\sqrt x$ is not integrable on $[1,\infty)$ (compare $\int_{[1,\infty)} f$ with the series $\sum 1/\sqrt n$). You can find a sequence of functions that converges to $f$ uniformly by truncating $f$: $$ f_n(x) = f(x)\chi_{[1,n]}(x), $$ $\chi_{[1,n]}(x)$ being the characteristic/indicator function of $[1,n]$.

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    $\begingroup$ Also $\int_1^x f(t)\,\mathrm dt=2(\sqrt x-1)$ so it clearly diverges. $\endgroup$
    – nejimban
    Commented Apr 19, 2022 at 17:42
  • $\begingroup$ Thanks for your answer. I think I understand the point about truncating the function. This ensures that $sup_{x \in [1,\infty)} |f(x) - f_n(x)|$ converges to $0$ since $\frac{1}{\sqrt{x}}$ is strictly decreasing, correct? However, I am struggling to understand your first sentence. Could you elaborate a bit more on how exactly I can compare that integral to the series? $\endgroup$
    – DerivativesGuy
    Commented Apr 19, 2022 at 18:11
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    $\begingroup$ @DerivativesGuy: Yes, since $f(x)\to 0$ as $x\to\infty$ is the key point. Actually, you can show $f(x)$ is not integrable by simply computing its integral using the functions $f_n(x)$ and the monotone convergence theorem since $0\le f_n\nearrow f$ pointwise. (This is essentially nejimban's comment.) But what I had in mind was simply bounding $f(x)$ on $x\in[n,n+1]$ from below by $1/\sqrt{n+1}$. Then it's clear $\int f\ge \sum1/\sqrt{n+1}$. $\endgroup$
    – Alex Ortiz
    Commented Apr 20, 2022 at 0:38
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    $\begingroup$ Thanks. I also figured it out myself after giving it another thought. You essentially define some simple functions $s_n = \sum \limits_{k=1}^{n} \frac{1}{\sqrt{k+1}} \chi_{k,k+1} \leq f$ whose integrals diverge to $\infty$. $\endgroup$
    – DerivativesGuy
    Commented Apr 20, 2022 at 6:46

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