Q) Let $\mu$ be a positive measure on $X$, $0<p<\infty, f\in L^p(\mu),f_n\in L^p(\mu), f_n(x)\to f(x)$ a.e. and $||f_n||_p\to ||f||_p$. Show that $\lim ||f_n-f||_p\to 0$ by completing the following $2$ steps:
Let me look at step $1$ for now.
$1$) By Egorov's theorem, $X=A\cup B$ in such a way that $\int_A |f|^p < \epsilon, \mu(B)<\infty$, and $f_n\to f$ uniformly on $B$. Apply Fatou's lemma to $\int_B|f_n|^p$ to show
$$\limsup \int_A|f_n|^pd\mu\leq \epsilon$$
Firstly, I know Egorov's theorem as: If $\mu(X)<\infty$, then $\exists A\subset X$, s.t. $\mu(A)<\epsilon$ and $f_n\to f$ uniformly on $X\setminus A$. So I don't understand how $\int_A |f|^p < \epsilon$ leads to a similar result?
Secondly, how can I apply Fatou's lemma to $\int_B|f_n|^p$ to get:
$$\limsup \int_A|f_n|^pd\mu\leq \epsilon$$
Best Answer
\begin{align*} \int_{B}\lim_{n}|f_{n}|^{p}&\leq\liminf_{n}\int_{B}|f_{n}|^{p}\\ &=\liminf_{n}\left(\int_{X}|f_{n}|^{p}-\int_{A}|f_{n}|^{p}\right)\\ &=\|f\|_{L^{p}}^{p}+\liminf_{n}\left(-\int_{A}|f_{n}|^{p}\right)\\ &=\|f\|_{L^{p}}^{p}-\limsup_{n}\int_{A}|f_{n}|^{p}, \end{align*} so \begin{align*} \limsup_{n}\int_{A}|f_{n}|^{p}\leq\|f\|_{L^{p}}^{p}-\int_{B}|f|^{p}=\int_{A}|f|^{p}<\epsilon. \end{align*}
For the first part:
Given $\epsilon>0$, since $|f|^{p}\in L^{1}$, by the absolute continuity of integral one has some $\delta>0$ such that \begin{align*} \int_{S}|f|^{p}<\epsilon/2,~~~~\mu(S)<\delta. \end{align*} Now it is clear that \begin{align*} \int|f|^{p}=\lim_{N}\int_{|f|^{p}\geq 1/2^{N}}|f|^{p}, \end{align*} then for some $N$ we have \begin{align*} \int_{|f|^{p}<1/2^{N}}|f|^{p}=\int|f|^{p}-\int_{|f|^{p}\geq 1/2^{N}}|f|^{p}<\epsilon/2. \end{align*} But we also have \begin{align*} \alpha^{p}\mu(|f|^{p}\geq\alpha)\leq\|f\|_{L^{p}}<\infty, \end{align*} in particular, we have $\mu(|f|^{p}\geq 1/2^{N})<\infty$. Applying Egorov to $(|f|^{p}\geq 1/2^{N})$ we have some $B\subseteq(|f|^{p}\geq 1/2^{N})$ such that \begin{align*} \mu((|f|^{p}\geq 1/2^{N})-B)<\delta \end{align*} and that $f_{n}\rightarrow f$ uniformly in $B$. Set $A=((|f|^{p}\geq 1/2^{N})-B)\cup(|f|^{p}<1/2^{N})$, then \begin{align*} \int_{A}|f|^{p}=\int_{(|f|^{p}\geq 1/2^{N})-B}|f|^{p}+\int_{|f|^{p}<1/2^{N}}|f|^{p}<\epsilon/2+\epsilon/2=\epsilon. \end{align*}
The proof of the so called absolute continuity:
We will prove that, for $g\in L^{1}$ and $\epsilon>0$, there exists some $\delta>0$ such that for every measurable set $S$ with $\mu(S)<\delta$, one has $\displaystyle\int_{S}|g|<\epsilon$. Then the above case is just a realization of $g$ to $|f|^{p}$.
Note that $|g|<\infty$ a.e. and hence $|g|\chi_{|g|\leq M}\uparrow|g|$ a.e. and hence \begin{align*} \int_{|g|>M}|g|=\int|g|-\int|g|\chi_{|g|\leq M}<\epsilon/2 \end{align*} for some $M>0$.
Then for any measurable set $S$ with $\mu(S)\leq\epsilon/(2M)$, we have \begin{align*} \int_{S}|g|&=\int_{S\cap(|g|\leq M)}|g|+\int_{S\cap(|g|>M)}|g|\\ &\leq M\int_{S\cap(|g|\leq M)}+\int_{|g|>M}|g|\\ &<M\int_{S}+\epsilon/2\\ &=M\mu(S)+\epsilon/2\\ &<\epsilon. \end{align*} The $\delta>0$ can be taken as $\epsilon/(2M)$.