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If ${f_i} \in L_{\rm loc}^1(\Omega )$ with $\Omega $ an open set in ${\mathbb R^n}$ , and ${f_i}$ are uniformly bounded in ${L^1}$ for every compact set, is it necessarily true that there is a subsequece of ${f_i}$ converging weakly to a regular Borel measure?

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Isn't the second condition the same as the first? – Jonas Teuwen Oct 10 '11 at 14:35
up vote 1 down vote accepted

Take $K_j$ a sequence of compact sets such that their interior grows to $\Omega$. That is, $\mathrm{int}(K_j) \uparrow \Omega$.

Let $f_i^0$ be a sub-sequence of $f_i$ such that $f_i^0|_{K_0}$ converges to a Borel measure $\mu_0$ over $K_0$.

For each $j > 0$, take a sub-sequence $f_i^j$ of $f_i^{j-1}$ converging to a Borel measure $\mu_j$ over $K_j$. It is evident, from the concept of convergence, that for $k \leq j$, and any Borel set $A \subset K_k$, $\mu_j(A) = \mu_k(A)$.

Now, define $\mu(A) = \lim \mu(A \cap K_j)$. And take the sequence $f_j^j$.

For any continuous $g$ with compact support $K$, there exists $k$ such that $K \subset \mathrm{int}(K_k)$ (why?). Then, since for $j \geq k$, $f_j^j|_{K_k}$ is a sequence that converges to $\mu_k$, $$ \int_{K_k} g f_j^j \mathrm{d}x \rightarrow \int_{K_k} g f_j^j \mathrm{d}\mu_k = \int g f_j^j \mathrm{d}\mu. $$

That is, $f_j^j \rightarrow \mu$.

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I have thought like this too, but how can you say μ is a measure? Moreover, how can you prove that μ is a regular Borel measure. – Hezudao Oct 13 '11 at 1:59
@Adterram: What is a regular measure? I guess that every measure over Borel sets of $\mathbb{R}^n$ is regular. – André Caldas Oct 13 '11 at 4:13
@Adterram: I guess the only trouble is showing that $\mu$ is $\sigma$-additive. Just take a sequence of disjoint measurable $A_n$. Now, $\mu(\bigcup_n A_n) = \lim_j \mu_j(K_j \cap \bigcup_n A_n) = \lim_j \sum_n \mu_j(A_n \cap K_j)$. Now, since we have only positive terms, we can exchange $\sum$ and $\lim$ (monotone convergence!) and get $\mu(\bigcup_n A_n) = \sum_n \lim_j \mu(K_j \cap A_n) = \sum_n \mu(A_n)$. – André Caldas Oct 13 '11 at 4:20

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