The question is Let $f_n \ge 0$ Be integrable and $f_n \rightarrow f$ a.e. Then $\lim_{n\rightarrow \infty} \mu(f_n)$ exists

My Proof: Firsly, we know that $f_n\rightarrow f$ And $f_n$ Is measurable , so is $f$. Then, $\inf_{m\ge n}f_m \le f_k$ for some $k\ge n$. Similarly, $\sup_{m\ge n}f_m \ge f_k$. Then, we know that $$\mu(\inf_{m\ge n}f_m)\le \mu ( f_k) \le \mu(\sup_{m\ge n}f_m)$$
Further, we know that $\lim_n \inf_{m\ge n} f_m \ge \inf_{m\ge n}f_m$, hence by monotone convergence theorem, we must have $\lim_n \mu(\inf f_n) = \mu(\lim \inf f_n)$. Similarly, we have $\lim_n \mu(\sup f_n) = \mu(\lim\sup f_n)$. But then, since $f_n \rightarrow f$, $\mu(\lim \inf f_n) = \mu (f) = \mu(\lim \sup f_n)$. Then, finally we have $$ \mu(f)=\mu(\inf_{m\ge n}f_m)\le \lim \mu ( f_k) \le \lim\mu(\sup_{m\ge n}f_m) = \mu(f)$$

Hence, $\lim \mu(f_n)$ exists

I think my proof is wrong, but I am not sure where...

  • $\begingroup$ What exactly is $\mu(f_n)$? $\endgroup$ – mechanodroid Sep 15 '17 at 21:23
  • $\begingroup$ @mechanodroid It is quite obvious from reading the question that it is the integral of $f_n$ with respect to the measure $\mu$. $\endgroup$ – uniquesolution Sep 15 '17 at 21:31
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    $\begingroup$ @uniquesolution It is not obvious at all and violates almost any customary notation in this setting. $\endgroup$ – zhw. Sep 15 '17 at 21:47
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    $\begingroup$ I agree with @zhw on this one. I've never seen $\mu(f)$ as a stand-in for $\int f\,\mathrm{d}\mu$. Even though I could certainly make a very good educated guess as to what $\mu(f)$ would refer to, there are ample notational choices that are more standard, any of which would be appropriate for asking a question here. $\endgroup$ – Michael L. Sep 15 '17 at 22:08
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    $\begingroup$ I have seen this notation in several papers and books, e.g. in Kallenberg's book. However, I would agree that this is not completely standard notation and it might have been helpful to explain it in the post. $\endgroup$ – Andre Sep 15 '17 at 22:18

I am assuming that $\mu(f) := \int f d\mu$. If that is what you mean, then I guess the problem would be in the phrase "Similarly, we have $\lim_n \mu(\sup f_n) = \mu(\lim\sup f_n)$". For example take the functions $f_n : [0,1]\rightarrow \mathbb{R}$ defined as \begin{equation*} \begin{cases} f_n(x) = n &\text{if } x<\frac{1}{n}\\ 0 &\text{otherwise} \end{cases} \end{equation*} For this functions we have $\mu(\sup_{m\geq n} f_m)\geq \mu(f_n) = 1$ and $\lim \sup f_n = 0$ so it is not true that $1\leq\lim_n \mu(\sup f_n) = \mu(\lim\sup f_n)=0$. Be careful that the sequence $\sup f_n$ is decreasing so I guess you'll have to replace the hypothesis that the sequence is positive in the monotone convergence theorem by "the sequence is bounded from above".

Moreover, what you claim in the title is false as far as I see. A counterexample can be made by taking the same sequence that I gave but instead of $f_n = n$ you should put a value for each $n$ that makes the integral take alternating values for each $n$. For example $n$ if $n$ is even and $2n$ if $n$ is odd. Hope this last part can be understood.


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