# Convergence question in measure theory

I have a convergence question in measure theory that requires assistance:

Let $1\leq p<\infty$. Suppose $f,\ f_n \in L^P$, and $f_n\to f$ in $L^P$. (i.e $(\int|f_n-f|^pd\mu)^{1\over p}\to 0$ as $n\to\infty$) Show that $\int|f_n|^pd\mu \to \int|f|^pd\mu$.

For $p=1$, $$\lvert\int|f_n|d\mu-\int|f|d\mu\rvert=\lvert\int(|f_n|-|f|)d\mu\rvert\leq\int\lvert|f_n|-|f|\rvert d\mu\leq\int|f_n-f|d\mu\to 0$$ hence obtaining the required inequality.

However, I have trouble with $p>1$.

Let us show it in two steps: $$f_n\to f\;\;\text{in } L^p\;\;\Longrightarrow\;\;|f_n|^p\to |f|^p \;\;\text{in }L^1\;\;\Longrightarrow \;\;\int |f_n|^p\,\mathrm d\mu\to\int |f|^p\,\mathrm d\mu.$$
The last implication should be pretty straightforward. For the first implication we might use that for $x,y\in\mathbb{R}$ one has $$||x|^p-|y|^p|\leq p(|x|^{p-1}+|y|^{p-1})|x-y|$$ and hence \begin{align} \int ||f_n|^p-|f|^p|\,\mathrm d\mu & \leq \int p(|f_n|^{p-1}+|f|^{p-1})|f_n-f|\,\mathrm d\mu \\ &=\int p|f_n|^{p-1}|f_n-f|\,\mathrm d\mu+\int p|f|^{p-1}|f_n-f|\,\mathrm d\mu\\ \end{align} which by the use of Hölder's inequality is less than or equal to $$p\left(\int |f_n|^{q(p-1)}\,\mathrm d\mu\right)^{1/q}\left(\int |f_n-f|^p\right)^{1/p}+p\left(\int |f|^{q(p-1)}\,\mathrm d\mu\right)^{1/q}\left(\int |f_n-f|\,\mathrm d\mu\right)^{1/p}\\$$ where $q>1$ is chosen such that $\frac1p +\frac1q=1$. This simplifies to (recall that $q(p-1)=p$) $$p\left[\left(\int |f_n|^p\,\mathrm d\mu\right)^{1/q}+\left(\int |f|^p\,\mathrm d\mu\right)^{1/q}\right]\left(\int |f_n-f|^p\,\mathrm d\mu\right)^{1/p}.$$ The last factor tends to $0$ as $n\to\infty$ by assumption, and thus we only need to argue that $$\sup_n\int |f_n|^p\,\mathrm d\mu<\infty.$$ This i'll leave for you to prove.
• It follows from the fact that $\int |f_n|^p\,\mathrm d\mu\leq \left[\left(\int |f-f_n|^p\,\mathrm d\mu\right)^{1/p}+\left(\int |f|^p\,\mathrm d\mu\right)^{1/p}\right]^p$ by Minkowski's inequality, where the first term goes to $0$ and second term is finite. Nov 20, 2013 at 18:06