# Fast convergence in $L^1$ implies convergence almost everywhere

This is a proof-verification request.

Claim: Let $(X,\mathscr M,\mu)$ be a measure space. Let $f_n$ ($n\in\mathbb N$) and $f$ be measurable, integrable, real-valued functions such that $(f_n)_{n\in\mathbb N}$ converges to $f$ in $L^1$ at a rate $O(1/n^p)$, where $p>1$. Then, $f_n\to f$ almost everywhere.

Note: If no assumption is made on the rate of convergence, then the best one can establish is the existence of a $\textit{sub}$sequence $(f_{n_k})_{k\in\mathbb N}$ converging to $f$ almost everywhere (Corollary 2.32 in Folland, 1999).

Proof of the Claim: Suppose there exists $M>0$ such that $\int|f_n-f|\,\mathrm d\mu\leq M/n^p$ for all $n\in\mathbb N$. For each $\varepsilon>0$ and $n\in\mathbb N$, let $$E(n,\varepsilon)\equiv \big\{x\in X\,\big|\,|f_n(x)-f(x)|\geq\varepsilon\big\}.$$ Then, $$\frac {M}{n^p}\geq\int|f_n-f|\,\mathrm d\mu\geq \int_{E(n,\varepsilon)}|f_n-f|\,\mathrm d\mu\geq\varepsilon\,\mu(E(n,\varepsilon))$$ for each $n\in\mathbb N$, so that $$\mu(E(n,\varepsilon))\leq\frac{M}{n^p\varepsilon}.$$ Defining $$E(\varepsilon)\equiv\bigcap_{m=1}^{\infty}\bigcup_{n=m}^{\infty}E(n,\varepsilon),$$ one has that $$\mu(E(\varepsilon))\underset{\forall m\in\mathbb N}{\leq}\mu\left(\bigcup_{n=m}^{\infty}E(n,\varepsilon)\right)\leq\sum_{n=m}^{\infty}\mu(E(n,\varepsilon))\leq\frac{M}{\varepsilon}\sum_{n=m}^{\infty}\frac{1}{n^p}\to0\quad\text{as m\to\infty},$$ since the series $\sum_{n=1}^{\infty}1/n^{p}=\zeta(p)$ converges. Therefore, $\mu(E(\varepsilon))=0$ for any $\varepsilon>0$, which implies also that $$\mu\left(\bigcup_{q\in\mathbb Q\cap(0,\infty)}E(q)\right)=0.$$

Now, if $x\in X$ is such that $f_n(x)\not\to f(x)$, then there exists some $q>0$ such that $q\in\mathbb Q$ and for each $m\in\mathbb N$, there exists some $n\geq m$ so that $|f_n(x)-f(x)|\geq q$. That is, $x\in E(q)$. Hence, the set where pointwise convergence fails is a subset of $\bigcup_{q\in\mathbb Q\cap(0,\infty)}E(q)$, completing the proof. $\quad\blacksquare$

• There is a typo in $E(\varepsilon)\equiv\bigcap_{m=1}^{\infty}\bigcup_{m=n}^{\infty}E(m,\varepsilon),$ since you've written $E(\varepsilon,m)$ earlier, i.e., the real argument comes first, plus you can't use the same variable of summation $m$ in two successive operators $\cap\cup$. In the next line $E(m,\varepsilon)$ looks like it ought to be $E(\varepsilon,n)$. – ForgotALot Dec 24 '15 at 2:49
• You could clarify where the quadratic convergence is actually needed. Any exponent p > 1 works just as fine. – shuhalo Dec 24 '15 at 3:18
• Your proof is fine. In fact, it can be easily generalized to the case $(f_n)_{n\in\mathbb N}$ converges to $f$ in $L^r$ at a rate $O(1/n^p)$, where $p>1$ and $r\geqslant 1$. – Ramiro Dec 24 '15 at 10:36
• Simpler proof: Monotone convergence shows $\int\sum|f_n-f| = \sum\int|f_n-f|<\infty$. Hence $\sum|f_n-f|<\infty$ almost everywhere, which implies $|f_n-f|\to0$ almost everywhere. – David C. Ullrich Dec 24 '15 at 16:05
• @triple_sec The $r$ in $L^r$ affects the rate at which $f_n \to f$ convergences in measure, because of the definition of the norm in $L^r$. Here are the details: If $(f_n)_{n\in\mathbb N}$ converges to $f$ in $L^r$ at a rate $O(1/n^p)$, where $p>1$ and $r\geqslant 1$, then $$\Vert f_n - f\Vert_r \leq \frac{M}{n^p}$$ which means $$\left (\int\vert f_n - f\vert ^r d\mu \right )^\frac{1}{r} \leq \frac{M}{n^p}$$ So we have $$\int\vert f_n - f\vert ^r d\mu \leq \frac{M^r}{n^{pr}}$$ So we get $$\mu(E(n,\varepsilon)) \leq \frac{M^r}{n^{pr} \varepsilon ^r}$$ – Ramiro Dec 25 '15 at 12:32

Moreover, instead of $$\frac{1}{n^p}$$ there could have been any function $$f$$, such that $$\Sigma_{i = 1}^\infty f(n)$$ converges.