I have the following theorem which I believe is true:
Suppose we have the measure space $(\mathscr{X},\mathcal{A}, \mu)$ and $p\in [1,\infty)$. Let $\{f_n\}_{n\in\mathbb{N}}\subseteq \mathcal{L}^p(\mathscr{X},\mathcal{A},\mu)$ and suppose that $f_n \overset{\mathcal{L}^p(\mu)}\longrightarrow f$ for some $f\in \mathcal{L}^0(\mathscr{X},\mathcal{A},\mu)$. Then
\begin{equation*}
\lim_{n\rightarrow\infty} \left\Vert f_n \right\Vert_p =\left\Vert f\right\Vert_p<\infty
\end{equation*}
The first equality is obvious because by the reverse triangle inequality, for each $n\in\mathbb{N}$, $\left\vert \left\Vert f_n\right\Vert_p – \left\Vert f\right\Vert_p\right\vert \le \left\Vert f_n – f\right\Vert_p$. Given the assumption that $f_n \overset{\mathcal{L}^p(\mu)}\longrightarrow f$, if we take the limit as $n\rightarrow\infty$ we may deduce that:
\begin{equation*}
\begin{split}
& \lim_{n\rightarrow\infty} \big\vert \left\Vert f_n\right\Vert_p – \left\Vert f\right\Vert_p\big\vert \le \lim_{n\rightarrow\infty}\left\Vert f_n – f\right\Vert_p =0\\
\Longrightarrow \qquad &\lim_{n\rightarrow\infty} \left(\left\Vert f_n\right\Vert_p – \left\Vert f\right\Vert_p\right) =0\\
\Longrightarrow \qquad &\lim_{n\rightarrow\infty} \left\Vert f_n\right\Vert_p = \left\Vert f\right\Vert_p
\end{split}
\end{equation*}
However, I just cannot see why it would be the case that $f\in \mathcal{L}^p(\mathscr{X},\mathcal{A},\mu)$ necessarily! I.e. I don't know why it is necessarily the case that $\left\Vert f \right\Vert_p<\infty$. Please help – thanks.
Best Answer
Because $\|f\|_{p}\leq\|f-f_{n_{0}}\|_{p}+\|f_{n_{0}}\|_{p}<1+\|f_{n_{0}}\|_{p}<\infty$ for some $n_{0}$.