The limit of locally integrable functions

Question

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?

Answer 1

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, $f_j^j \rightarrow \mu$.