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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. $\quad\blacksquare$

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  • $\begingroup$ 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 $E(\varepsilon,n)$. $\endgroup$
    – ForgotALot
    Commented Dec 24, 2015 at 2:49
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    $\begingroup$ You could clarify where the quadratic convergence is actually needed. Any exponent p > 1 works just as fine. $\endgroup$
    – shuhalo
    Commented Dec 24, 2015 at 3:18
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    $\begingroup$ 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 $r\geqslant 1$. $\endgroup$
    – Ramiro
    Commented Dec 24, 2015 at 10:36
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    $\begingroup$ 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. $\endgroup$ Commented Dec 24, 2015 at 16:05
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    $\begingroup$ @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 $$ \mu(E(n,\varepsilon)) \leq \frac{M^r}{n^{pr} \varepsilon ^r}$$ $\endgroup$
    – Ramiro
    Commented Dec 25, 2015 at 12:32

3 Answers 3

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Yes, your proof is correct.

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

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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Jul 31, 2023 at 8:38
  • $\begingroup$ @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. $\endgroup$
    – shark
    Commented Aug 12, 2023 at 14:42
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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.

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