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To build a bridge from the local approximation to the global approximation, the following argument in measure theory is usually used in PDE, with which I don't feel very comfortable.

Suppose $U$ is a nonempty, bounded open subset of $\mathbb{R}^n$. Suppose $f$ is locally integrable in $U$ such that $\sup_{V\Subset U}\|f\|_{L^1(V)}<\infty$. Then $$ \sup_{V\Subset U}\|f\|_{L^1(V)}=\|f\|_{L^1(U)}, $$ where $V\Subset U$ means $V\subset\overline{V}\subset U$ and $\overline{V}$ is compact.

It is trivial that one has the inequality $$ \sup_{V\Subset U}\|f\|_{L^1(V)}\leq\|f\|_{L^1(U)}. $$

Could anyone give the other direction?

2 Answers2

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You may find an increasing sequence of compact subsets $K_n$, for example, $K_n =\{ x\in U: d(x,U^c)\geq \frac{1}{n}\}$ so that $\bigcup_n K_n=U$. Then by monotone convergence (${\bf 1}_K$ being an indicator function): $$ \lim_n \|f\|_{L^1(K_n)} = \lim_n \int_U |f| \; {\bf 1}_{K_n} \; dx = \int_U |f| \; dx = \|f\|_{L^1(U)}\in [0,+\infty]$$ If you want a sup over open subsets you may simply take $V_n =\{ x\in U: d(x,U^c)>\frac{1}{n}\}$

H. H. Rugh
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You are trying to establish that if the $\sup$ is bounded, then $f \in L^1(U)$.

Let $U_n = \{x \in U | B(x,{1 \over n}) \subset U \}$. We see that $\cup_n U_n = U$, and $\overline{U}_n \subset U$. Let $f_n = f \cdot 1_{U_n}$, and we have $|f_n(x)| \le |f_{n+1}(x)|$ and $|f_n(x)| \to |f(x)|$ for all $x \in U$. The monotone convergence theorem shows that $\int_U |f_n| \to \int_U |f|$.

Hence if $\sup_{V\Subset U}\|f\|_{L^1(V)}<\infty$, then $\int_U |f|$ is bounded and hence $f \in L^1(U)$ and $\sup_{V\Subset U}\|f\|_{L^1(V)} = \|f\|_{L^1(U)}$.

copper.hat
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