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 have verified that $(\tilde L^p (\mathbb R^d), \|\cdot\|_{\tilde L^p})$ is complete. Let $(x_m)$ be a countable dense subset of $\mathbb R^d$. The sphere has Lebesgue measure $0$. Then by dominated convergence theorem (DCT),
$$
\|f\|_{\tilde L^p} = \sup_{m \in \mathbb N} \|1_{B(x_m, 1)} f\|_{L^p}
\quad \forall f \in \tilde L^p (\mathbb R^d).
$$
I would like to ask if a version of DCT holds for $L^p (\mathbb R^d)$, i.e.,
Let $f_n, g \in \tilde L^p (\mathbb R^d)$ such that $|f_n| \le g$ a.e. for all $n$. If there is $f:\mathbb R^d \to \mathbb R$ such that $f_n \to f$ a.e., then $f \in \tilde L^p (\mathbb R^d)$ and $\|f_n-f\|_{\tilde L^p} \to 0$.
Thank you so much for your elaboration!
Best Answer
A counterexample is given by $f_n=1_{(n,\infty)}$, $g=1$, and $f=0$.
Then all the conditions on $f_n,g,f$ hold, but $\|f_n-f\|_{\tilde L^p} \not\to0$.