Limit of the p-norm of a function on subdomains equals the p-norm of the function on the union domain Let $\Omega$ be an open subset in $\mathbb{R}^n$. Given a measurable function $f$, define
$$
||f||_{p,\Omega}=\inf_{a\in\mathbb{R}}||f-a||_{L^p{(\Omega})}.
$$
Let $\{\Omega_n\}$ be a sequence of open subsets such that $\Omega_n\subseteq \Omega_{n+1}$ for all $n$. Prove that
$$
\lim_{n\to\infty}||f||_{p,\Omega_n}=||f||_{p,\cup_{n=1}^\infty\Omega_n}.
$$
My attempt:
for any $a$, by levi's monotone theorem, I obtained
$$
\lim_{n\to\infty}||f-a||_{L^p(\Omega_n)}=||f-a||_{L^p(\cup_{n=1}^\infty \Omega_n)}.
$$
But this only implies
$$
\limsup_{n\to\infty}||f||_{p,\Omega_n}\leq ||f||_{p,\cup_{n=1}^\infty\Omega_n}....
$$
How to prove the other direction?
 A: First, let $a\in\mathbb{R}$ be arbitrary. As you noted, this implies
$$
\limsup_{n\rightarrow\infty}\left\Vert f\right\Vert _{p,\Omega_{n}}\leq\lim_{n\rightarrow\infty}\left\Vert f-a\right\Vert _{L^{p}\left(\Omega_{n}\right)}=\left\Vert f-a\right\Vert _{L^{p}\left(\bigcup\Omega_{n}\right)},
$$
and thus
$$
\limsup_{n\rightarrow\infty}\left\Vert f\right\Vert _{p,\Omega_{n}}\leq\left\Vert f\right\Vert _{p,\bigcup\Omega_{n}}.
$$
In the case $\limsup_{n\rightarrow\infty}\left\Vert f\right\Vert _{p,\Omega_{n}}=+\infty$,
this implies the claim, so that we can assume $\limsup_{n\rightarrow\infty}\left\Vert f\right\Vert _{p,\Omega_{n}}<\infty$.
The rest is basically an exercise in the direct method of the calculus of variations (if you want to put it fancy). We take a sequence $(a_n)_n$ of minimizers, extract a convergent subsequence and show that the limit does what we want:
Choose a sequence $\left(a_{n}\right)_{n\in\mathbb{N}}$ with
$$
\left\Vert f\right\Vert _{p,\Omega_{n}}\leq\left\Vert f-a_{n}\right\Vert _{L^{p}\left(\Omega_{n}\right)}\leq\left\Vert f\right\Vert _{p,\Omega_{n}}+\frac{1}{n}.
$$
If $\limsup_{n\rightarrow\infty}\left|a_{n}\right|=\infty$, then
there is a subsequence $\left(a_{n_{k}}\right)$ with $\left|a_{n_{k}}\right|\xrightarrow[k\rightarrow\infty]{}\infty$,
which implies
\begin{eqnarray*}
\left\Vert f-a_{n_{k}}\right\Vert _{L^{p}\left(\Omega_{n_{k}}\right)} & \geq & \left\Vert f-a_{n_{k}}\right\Vert _{L^{p}\left(\Omega_{n_{k_{0}}}\right)}\\
 & \geq & \left\Vert a_{n_{k}}\right\Vert _{L^{p}\left(\Omega_{n_{k_{0}}}\right)}-\left\Vert f-a_{n_{k_{0}}}\right\Vert _{L^{p}\left(\Omega_{n_{k_{0}}}\right)}\\
 & \geq & \left|a_{n_{k}}\right|\cdot\lambda\left(\Omega_{n_{k_{0}}}\right)-\left[\left\Vert f\right\Vert _{p,\Omega_{n_{k_{0}}}}+\frac{1}{n_{k_{0}}}\right]\\
 & \xrightarrow[k\rightarrow\infty]{} & \infty.
\end{eqnarray*}
Here, we have chosen $k_{0}\in\mathbb{N}$ with the property $\lambda\left(\Omega_{n_{k_{0}}}\right)>0$
(where $\lambda$ denotes Lebesgue measure) and $\left\Vert f\right\Vert _{p,\Omega_{n_{k_{0}}}}<\infty$.
Such a $k_{0}$ exists because of $\limsup_{n\rightarrow\infty}\left\Vert f\right\Vert _{p,\Omega_{n}}$
and because $\bigcup\Omega_{n}$ is not a null-set (otherwise, the
claim would be trivial).
Thus, $\left(a_{n}\right)_{n\in\mathbb{N}}$ is bounded. Let $\left(n_{k}\right)_{k\in\mathbb{N}}$
with $\left\Vert f\right\Vert _{p,\Omega_{n_{k}}}\xrightarrow[k\rightarrow\infty]{}\liminf_{n\rightarrow\infty}\left\Vert f\right\Vert _{p,\Omega_{n}}$
and choose (Bolzano-Weierstraß) a convergent subsequence $\left(a_{n_{k_{\ell}}}\right)_{\ell}$
with $a_{n_{k_{\ell}}}\xrightarrow[\ell\rightarrow\infty]{}a\in\mathbb{R}$.
Using Fatou's Lemma, we get
\begin{eqnarray*}
\left\Vert f\right\Vert _{p,\bigcup\Omega_{n}}^{p} & \leq & \left\Vert f-a\right\Vert _{p,\bigcup\Omega_{n}}^{p}\\
 & = & \int_{\bigcup\Omega_{n}}\liminf_{\ell\rightarrow\infty}\left[\chi_{\Omega_{n_{k_{\ell}}}}\cdot\left|f-a_{n_{k_{\ell}}}\right|^{p}\right]\, dx\\
 & \leq & \liminf_{\ell\rightarrow\infty}\int_{\bigcup\Omega_{n}}\chi_{\Omega_{n_{k_{\ell}}}}\cdot\left|f-a_{n_{k_{\ell}}}\right|^{p}\, dx\\
 & = & \liminf_{\ell\rightarrow\infty}\left\Vert f-a_{n_{k_{\ell}}}\right\Vert _{L^{p}\left(\Omega_{n_{k_{\ell}}}\right)}\\
 & \leq & \liminf_{\ell\rightarrow\infty}\left[\left\Vert f\right\Vert _{p,\Omega_{n_{k_{\ell}}}}+\frac{1}{n_{k_{\ell}}}\right]\\
 & = & \lim_{k\rightarrow\infty}\left\Vert f\right\Vert _{p,\Omega_{n_{k}}}\\
 & = & \liminf_{n\rightarrow\infty}\left\Vert f\right\Vert _{p,\Omega_{n}}\\
 & \leq & \limsup_{n\rightarrow\infty}\left\Vert f\right\Vert _{p,\Omega_{n}}\\
 & \leq & \left\Vert f\right\Vert _{p,\bigcup\Omega_{n}},
\end{eqnarray*}
which completes the proof.
