Convergence in $L^p$ with given properties

Question

Let $p\in [1,\infty[$, $(\Omega, \mathcal A, \mu)$ be a probability space and $(f_n)_{n\in \mathbb N}$ be a sequence of measurable functions $f_n : \Omega \to \mathbb R$ and $f:\Omega \to \mathbb R$ with the following properties:

$\forall \omega \in \Omega: f_n(\omega) \to f(\omega)$ and
$\sup_{n\in \mathbb N}\left \lVert f_n 1_{\{\lvert f_n \rvert > m\}}\right \rVert_p \to 0$ as $m\to \infty.$

Show that $$\lVert f_n - f \rVert_p \to 0.$$

I was able to prove one inequality:
We have $$0 = \int \lvert f-f|^p d\mu = \int \liminf_{n\to \infty} |f_n - f|^p d\mu \leq \liminf_{n\to \infty}\int \lvert f_n - f \rvert ^p d\mu = \liminf_{n\to \infty} \lVert f_n -f\rVert_p^p$$ by Fatous lemma. But how to prove the other direction? The second property is hard for me to understand. I tried to get a bound involving the measure of a set since $\mu $ is finite but feel like I need a hint.

Answer 1

I'll take $p=1$ for simplicity. Claim: $\{f_n\}$ is uniformly integrable.

Proof: Let $\epsilon>0.$ Choose $m$ such that $\int_{|f_n|>m} |f_n| <\epsilon/2$ for all $n.$ Set $\delta = \epsilon/(2m).$ If $\mu(E) < \delta,$ then

$$\int_E |f_n| = \int_{E\cap \{|f_n|\le m\}} |f_n| + \int_{E\cap \{|f_n|> m\}} |f_n| \le m\cdot \delta + \epsilon/2 <\epsilon.$$

This proves the claim. Fatou's lemma then makes it clear that $\{f_n\}\cup \{f\}$ is uniformly integrable. Hence $\{ |f_n-f|\}$ is uniformly integrable.

Let $\epsilon>0$ again. Choose $\delta >0$ such that $\mu(E) < \delta$ implies $\int_E |f_n-f| < \epsilon$ for all $n.$ By Egorov, there exists a set $E, \mu(E) < \delta,$ such that $f_n\to f$ uniformly on $\Omega \setminus E.$ Thus

$$\int_{\Omega} |f_n-f| \le \int_{\Omega \setminus E} |f_n-f| + \int_E |f_n-f| < \epsilon.$$

The first integral on the right $\to 0$ by uniform convergence and the fact that we're on a finite measure space. Thus $\limsup \int_{\Omega} |f_n-f| \le \epsilon.$ Since $\epsilon$ was arbitrary, we have $\int_{\Omega} |f_n-f| \to 0$ as desired.

Answer 2

Suppose $M_0 \ge 0$ such that $\sup_{n\in \mathbb N}\left \lVert f_n 1_{\{\lvert f_n \rvert > M_0\}}\right \rVert_p \le 1$.

• $(||f_n||_p)_n$ is uniformly bounded. $$\sup_{n\in \mathbb N}\left \lVert f_n \right \rVert_p \le \sup_{n\in \mathbb N}\left \lVert f_n 1_{\{\lvert f_n \rvert \le M\}}\right \rVert_p + \sup_{n\in \mathbb N} \left \lVert f_n 1_{\{\lvert f_n \rvert > M\}}\right \rVert_p \le M_0 + 1$$
• $f \in L^p(\Omega, \mathcal A, \mu)$: due to pointwise convergence $f_n \to f$, $|f| = \lim\limits_{n\to\infty}|f_n|$, so $|f| = \limsup_n |f| \le \sup_n |f_n|$. $$||f||_p \le \sup ||f_n||_p \le M_0 + 1$$
• $f_n \to f$ in $L^p(\Omega, \mathcal A, \mu)$: For any $M>0$, \begin{align} & ||f_n - f||_p \\ &\le ||(f_n - f) 1_{\{|f_n-f| \le \epsilon\}}||_p + ||(f_n - f) 1_{\{\epsilon < |f_n-f| \le 2M\}}||_p + ||(f_n - f) 1_{\{|f_n-f| > 2M\}}||_p \\ &\le \epsilon + 2M \mu \{|f_n-f|>\epsilon\} + || (|f_n| + |f|) 1_{\{|f_n| \vee |f| > M\}} ||_p \\ &\le \epsilon + 2M \mu \{|f_n-f|>\epsilon\} + 2 \underbrace{|||f_n| 1_{\{|f_n|>M\}}||_p}_{|f_n| \text{ is larger}} + 2 \underbrace{|||f| 1_{\{|f|>M\}}||_p}_{|f| \text{ is larger}} \\ &\le \epsilon + 2 \underbrace{M\mu \{|f_n-f|>\epsilon\}}_{\substack{\text{convergence a.e implies} \\ \text{convergence in measure} \\ \text{in probability space}}} + 2 \underbrace{\sup_{n\in \mathbb N}\left \lVert f_n 1_{\{\lvert f_n \rvert > M \}}\right \rVert_p }_{\substack{\text{given: no escape from} \\ \text{vertical infinity}}} + 2 \underbrace{|||f| 1_{\{|f|>M\}}||_p}_{f \in L^p(\Omega,\cal A, \mu)} \end{align}

Let $M > 0$ be sufficiently large so that

1. $\sup_{n\in \mathbb N}\left \lVert |f_n| 1_{\{\lvert f_n \rvert > M \}}\right \rVert_p < \epsilon$
2. $|||f| 1_{\{|f|>M\}}||_p < \epsilon$

Let $N \in \Bbb{N}$ be sufficiently large so that for all $n \ge N$, $\mu\{|f_n-f|>\epsilon\} \le \dfrac{\epsilon}{M}$

Hence, $||f_n - f||_p \le 7 \epsilon$ for all $n \ge N$. This shows $f_n \xrightarrow[n\to\infty]{L^p(\Omega,\cal A, \mu)} f$

Remarks:

1. Terence Tao describes the second property in the question body as "no escape to vertical infinity".
2. I've skipped one step to simplify writings of indicator functions. Please refer to my discussions with saz in the comments below.
3. We have actually proven the result with a weaker assumpion: convergence in measure $f_n \stackrel{\mu}{\to} f$ instead of convergence almost everywhere, based on the assumption that $\mu(\Omega) < +\infty$.
   • The advantage of this approach is that we don't need almost everywhere convergence, which is necessary in order to apply Egorov's theorem.
   • Dropping the condition $\mu(\Omega) < +\infty$ invalidates this answer as the RHS of $$||(f_n - f) 1_{\{|f_n-f| \le \epsilon\}}||_p \le \epsilon \mu(\Omega)$$ is no longer finite.