Fix $p \in [1, \infty)$. Let $(L^p (\mathbb R^d), \|\cdot\|_{L^p})$ be the Lesbesgue space of $p$-integrable real-valued functions on $\mathbb R^d$. Let $\tilde L^p (\mathbb R^d)$ be the space of Lebesgue measurable functions $f:\mathbb R^d \to \mathbb R$ such that $$ \|f\|_{\tilde L^p} := \sup_{x \in \mathbb R^d} \|1_{B(x, 1)} f\|_{L^p} < \infty, $$ where $B(x, 1)$ is the open unit ball centered at $x$. I would like to generalize this result to higher dimensions, i.e.,
$(\tilde L^p (\mathbb R^d), \|\cdot\|_{\tilde L^p})$ is complete.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?
Let $(x_m)$ be a countable dense subset of $\mathbb R^d$. By dominated convergence theorem, $$ \|f\|_{\tilde L^p} := \sup_{m \in \mathbb N} \|1_{B(x_m, 1)} f\|_{L^p} \quad \forall f \in \tilde L^p (\mathbb R^d). $$
Let $(f_n)$ be a Cauchy sequence in $\tilde L^p (\mathbb R^d)$. For each $m \ge 1$, there is $g_m \in L^p (\mathbb R^d)$ such that $\|1_{B(x_m, 1)} (f_n-g_m)\|_{L^p} \xrightarrow{n \to \infty} 0$. By diagonal extraction, there is a subsequence $\varphi$ of $\mathbb N$ and a (Lebesgue) null subset $N$ of $\mathbb R^d$ such that $f_{\varphi (n)} \xrightarrow{n \to \infty} g_m$ pointwise on $B(x_m, 1) \setminus N$. Then $g_m =g_n$ on $(B(x_m, 1) \setminus N) \cap (B(x_n, 1) \setminus N)$. Let $$ g := g_m \sum_m 1_{B(x_m, 1) \setminus N}. $$
We will prove that $\|f_n -g\|_{\tilde L^p} \to 0$. Fix $\varepsilon >0$, there is $N >0$ such that $$ \sup_{m \in \mathbb N} \|1_{B(x_m, 1)} (f_n - f_{n'})\|_{L^p} < \varepsilon \quad \forall n, n' > N. $$
Then $$ \|1_{B(x_m, 1)} (f_n - f_{n'})\|_{L^p} < \varepsilon \quad \forall n, n' > N, m \in \mathbb N. $$
Then $$ \|1_{B(x_m, 1)} (f_n - g)\|_{L^p} < \varepsilon \quad \forall n>N, m \in \mathbb N $$
Then $$ \sup_{m \in \mathbb N} \|1_{B(x_m, 1)} (f_n - g)\|_{L^p} \le \varepsilon \quad \forall n>N. $$
This completes the proof.
