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This question has been already asked on this site before here - Proving Fatou type lemma I found that the solution provided is quite sophisticated. There was a suggestion in the comments section to first prove it for intervals and then for a countable union of intervals. I was also thinking along the same lines. I have shown that the required inequality is actually an equality if $U$ is a connected open subset of $\mathbb{R}$ or if $U$ is a finite union of disjoint connected open subsets of $\mathbb{R}$. But I am unable to generalize it for countable unions. I thought of using Fatou's Lemma, but to no avail. So can someone please provide me a proof using basic measure theory,i.e using convergence theorems, regularity properties, etc.? Thanks for any help.

Ester
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I guess you can proof it by using standard measure theory:

First define:

$$\mu_n(U) := \int_U f_n(x)dx$$

And $$\mu(U) := \lim_{n\to\infty} \mu_n(U)$$

Then $\mu$ is a $\sigma$-finite measure on $\mathcal{B}(\Bbb R)$

On the other hand you have with $$\psi(U) :=\int_U f(x)dx$$ a second $\sigma$-finite measure on $\mathcal{B}(\Bbb R)$.

Now consider $$\mathcal{E} = \left\{ (-\infty,y] \;|\; y \in \Bbb R\right\}$$ which is an intersection-stable generator of $\mathcal{B}(\Bbb R)$

But by your given properties it holds $$\mu|_\mathcal{E} = \psi|_\mathcal{E}$$

And so $\mu \equiv \psi$ on $\mathcal{B}(\Bbb R)$, hence especially for open sets $U \in \mathcal{B}(\Bbb R)$

Gono
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  • Sorry, I don't quite understand why mu is a measure? Can you please explain? – Ester May 08 '17 at 11:34
  • Because it fulfill all properties of a meaure…$$\mu(\emptyset) = \lim_{n\to\infty} \mu_n(\emptyset) = 0$$ and also $\sigma$-additivity holds by $$\begin{align} \mu\left(\bigcup_{k=1}^\infty A_k\right) &= \lim_{n\to\infty} \mu_n \left(\bigcup_{k=1}^\infty A_k\right) \ &= \lim_{n\to\infty} \sum_{k=1}^\infty \mu_n(A_k) \ &= \sum_{k=1}^\infty \lim_{n\to\infty} \mu_n(A_k) \ &= \sum_{k=1}^\infty \mu(A_k) \end{align} $$ – Gono May 08 '17 at 12:49
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    I hope you are aware, that the $\mu_n$ are measures too… that's what I used in my proof in the former comment. You can show this by using properties of the integral of just check https://en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem – Gono May 08 '17 at 12:58
  • Yes, I know that mu_n are measures , but how does the limit goes inside the summation in your second last step. – Ester May 08 '17 at 13:11
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    Indeed this is a non-trivial-step… thought first it can be shown by some monotone convergence or dominated convergence argument, but it doesn't seem to work that way. For a detailed proof e.g. check https://math.stackexchange.com/a/120219/384471

    Otherwise… if you also have that $\max{f_1, f_2, \ldots}$ is integrable or that the limit of the $\mu_n$ is monotone you could use dominated or monotone convergence above.

     – Gono May 08 '17 at 13:59
  • Why is it true that lim mu_(n)(E) exists for every measurable subset E of R?By hypothesis, we can only say that it exists for connected open subsets of R or finite union of them. It seems that the proof goes through only after we prove the existence of this limit for every measurable subset E of R. – Ester May 08 '17 at 16:49
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You may use Borel regularity of the measure $d\mu = f\, dx$.

Given an open $U$ and $\epsilon>0$ there is a compact set $K\subset U$ with $\mu(U\setminus K)<\epsilon$. By compactness there are finitely many open disjoint intervals $I_1,...,I_k$ with $K\subset S=I_1\cup...\cup I_k\subset U$. Now, with $\mu_n = f_n \, dx$ we have $\mu_n(S) \rightarrow \mu(S)$ so $$\liminf_n \;\mu_n(U)\;\geq \;\lim_n \mu_n(S) = \mu(S) \geq \mu(U)-\epsilon.$$

H. H. Rugh
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