# 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 $E(\varepsilon,n)$. Commented Dec 24, 2015 at 2:49
• You could clarify where the quadratic convergence is actually needed. Any exponent p > 1 works just as fine. Commented Dec 24, 2015 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 $r\geqslant 1$. Commented Dec 24, 2015 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. Commented Dec 24, 2015 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 $$\mu(E(n,\varepsilon)) \leq \frac{M^r}{n^{pr} \varepsilon ^r}$$ Commented Dec 25, 2015 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.

I'll refer to the corresponding exercise and notations in Tao's "Introduction to Measure Theory".

Given a sequence $${f_n = A_n 1_{E_n}}$$ of step functions, we have the $${N^{th}}$$ tail support $${E^*_N := \bigcup_{n \geq N} E_n}$$ of the sequence $${f_1, f_2, f_3, \ldots}$$

Exercise $$1.5.5$$ (Fast $${L^1}$$ convergence) Suppose that $${f_n, f: X \rightarrow {\bf C}}$$ are measurable functions such that $${\sum_{n=1}^\infty \|f_n-f\|_{L^1(\mu)} < \infty}$$.

Show that $${f_n}$$ converges pointwise almost everywhere to $${f}$$.

Proof:

We first work in the special case where $${f_n = A_n 1_{E_n}}$$ are step functions of measurable sets $${E_n}$$ and $$f = 0$$. For simplicity we will assume that the $${A_n > 0}$$, and that $${\mu(E_n) > 0}$$. We also assume that either the $${A_n}$$ converge to zero, or else they are bounded away from zero.

From condition we have $$\sum_{n=1}^\infty \int_{X} A_n 1_{E_n}(x) d\mu = \sum_{n=1}^\infty A_n\mu(E_n) < \infty$$. Suppose that the $$A_n$$'s are bounded away from $$0$$. (i.e. there exists $${c>0}$$ such that $${A_n \geq c}$$ for every $${n}$$). Then clearly $$\sum_{n=1}^\infty \mu(E_n) < \infty$$. By the Borel-Cantelli lemma, a.e $$x \in X$$ is in at most finitely many $$E_n$$. Note that $$\forall x \in {\bigcap_{N=1}^\infty E^*_N}$$, $$x \in E_n$$ for infinitely many $$n$$. Indeed, if $$x \in E_n$$ for finitely many $$n$$, pick the largest $$n$$ s.t $$x \in E_n$$. But $$x \in E^*_N$$ for any $$N > n$$ implies that $$n$$ is not the largest index, a contradiction. Hence $$\mu({\bigcap_{N=1}^\infty E^*_N}) = 0$$ and $$f_n \rightarrow 0$$ p.w a.e by (v) of Exercise $$1.5.3$$.

Now for general measurable $${f_n, f: X \rightarrow {\bf C}}$$. Suppose for contradiction that $$\exists \varepsilon > 0$$ such that the set $$\{x \in X: \lim_{n \to \infty}|f_n(x) - f(x)| > \varepsilon \} = A$$ has positive measure. Let $$E_n := \{x \in X: |f_n(x) - f(x)| > \varepsilon\}$$, then $$E_n$$ is measurable and $$A = {\bigcap_{N=1}^\infty E^*_N}$$. Since the function $$\varepsilon 1_{E_n} \leq |f - f_n|$$ for each $$n$$, we have: $$\sum_{n=1}^\infty \int_X \varepsilon 1_{E_n} d\mu \leq \sum_{n=1}^\infty \int_X|f_n-f|d\mu < \infty$$. By the special case above, we thus have $$\mu({\bigcap_{N=1}^\infty E^*_N}) = \mu(A) = 0$$, a contradiction. The claim then follows by setting $${\varepsilon = 1/m}$$ for $${m=1,2,3,\ldots}$$ and using the fact that the countable union of null sets is again a null set.

• I'm not actually understanding the assumption $A_n$ is bounded away from zero by some constant c .I know if the sequence does not converge to zero, then there is a subsequent with the property you mentioned but yet you're inequalities yet need to be verified. Commented Jul 31, 2023 at 8:38
• @KhaledAlekasir: The above argument runs along the said subsequence just as well, see Section $1.5.2$ of 'An Introduction to Measure Theory' by Tao. Commented Aug 12, 2023 at 14:42

for each $$q \in \mathbb{Q^+}$$ consider $$\{E_{i}^{q}\}_{i=1}^{\infty}$$ such that $$E_{i}^{q} = \{x \in X: |f_n(x) - f(x)| \geq q\}$$ by Chebyshev's inequality $$\mu(E_{i}^{q}) \leq \frac{|f_n-f|_{L^1}}{q}$$ and since q is fixed $$\Sigma_{i=0}^{\infty}\mu(E_{i}^{q}) < \infty$$ using Borell-Cantli's lemma then $$\mu(\limsup E_{i}^{q}) = 0$$ for any $$q \in \mathbb{Q}^+$$.
By using countable additivity, conclusion follows.