Show that $\lim_{n\to\infty}\|f_n\|_1=0$ under some reasonable conditions

measure-theoryreal-analysissolution-verification

I would be glad if someone could check my work about the following exercise. Also, I couldn't explain the equality $(*)$ which is interchanging the limits. Probably, I should use uniform integrability at some point. Thanks!

Let $(f_n)_{n=1}^\infty$ be sequence of measurable functions on a finite measure space $(X,\mathcal M, \mu)$. Recall that $(f_n)_{n=1}^\infty$ is said to be uniformly integrable if for every $\epsilon>0$ there exists $\delta>0$ such that $\int_E|f_n|<\epsilon$ for all measurable sets $E\subseteq X$ satisfying $\mu(E)<\delta$ and all $n\in\Bbb N$. Prove that if $(f_n)_{n=1}^\infty$ is uniformly integrable, $sup_{n}\|f_n\|_1<\infty$, and $(f_n)_{n=1}^\infty$ converges in measure to $0$, then $\lim_{n\to\infty}\|f_n\|_1=0$.

$Attempt.$ Since $(f_n)_{n=1}^\infty$ converges in measure to $0$, we have
$$\lim_{n\to\infty}\mu(\{x\in X: |f_n(x)|>\eta\})=0\qquad \text{for all $\eta >0$}.$$
In particular, $\lim_{n\to\infty}\mu(E_{k,n})=0$ for all $k\in\Bbb N$, where $E_{k,n}=\{x\in X: |f_n(x)|>\frac{1}{k}\}$ for $n\in \Bbb N$. Note that for any $n\in \Bbb N$ we have $E_{1,n}\subseteq E_{2,n} \subseteq\dots$ and $X=E_n\sqcup \bigcup_{k=1}^\infty E_{k,n}$ where $E_n=\{x\in X: |f_n(x)|=0\}$, so $\int_{E_{1,n}}|f_n|\leq \int_{E_{2,n}}|f_n|\leq\dots$ and $\int_X|f_n|=\int_{E_n}|f_n|+\int_{\cup_{k=1}^\infty E_{k,n}}|f_n|=\lim_{k\to\infty}\int_{E_{k,n}}|f_n|$. So we get $\|f_n\|_1=\lim_{k\to\infty}\int_{E_{k,n}}|f_n|$. So

\begin{align*}0\leq\lim_{n\to\infty}\|f_n\|_1=\lim_{n\to\infty}\left(\lim_{k\to\infty}\int_{E_{k,n}}|f_n|\right) &\overset{!}{\leq} \lim_{n\to\infty}\left(\lim_{k\to\infty}\int_{E_k,n}Md\mu\right)\qquad \text{for some $M\geq 0$ since $\sup_{n}\|f_n\|_1<\infty$.}
\\
&=\lim_{n\to\infty}\left(\lim_{k\to\infty}M\mu(E_{k,n})\right)
\\
&=M\lim_{n\to\infty}\lim_{k\to\infty}\mu(E_{k,n})
\\
&\overset{(*)}{=}M\lim_{k\to\infty}\lim_{n\to\infty}\mu(E_{k,n})
\\
&=M\lim_{k\to\infty}(0)=M.0=0
\end{align*}

So, $\lim_{n\to\infty}\|f_n\|_1=0$.

Best Answer

Here is a hint that may help you get started. A common theme in measure theory problems is to truncate our sequence $(f_n)$. Let $\varepsilon > 0$ be given, and write $$ \int_X |f_n| = \int_{|f_n| < \varepsilon}|f_n| + \int_{|f_n|\ge \varepsilon}|f_n|. $$ Try thinking about how to make both terms smaller than a constant times $\varepsilon$ by taking $n$ sufficiently large.

Two additional points:

  1. The result is not true on an infinite measure space. Consider the sequence $f_n = (2n)^{-1}1_{[-n,n]}$ on $\mathbf R$ with Lebesgue measure.
  2. The assumption that $\sup_n\|f_n\|_1 < \infty$ is not needed, and is in fact a consequence of the uniform integrability of the sequence $(f_n)$ and the convergence to $0$ in measure. The proof of this fact is very similar to the problem you are trying to solve.
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