# Two questions regarding convergence almost everywhere:

For the purpose of this post $\mathcal{L}$ denotes the Lebesgue $\sigma$ -algebra, $\mathcal{B}$ denotes the Borel $\sigma$ -algebra and $\lambda$ denotes the Lebesgue Measure.

I'm trying prove the following two claims:

1. Let $A\subseteq\mathbb{R}$ be Lebesgue measurable with $\lambda\left(A\right)=\infty$ and let $f_{n}:\left(A,\mathcal{L}\right)\to\left(\mathbb{R},\mathcal{B}\right)$ be a sequence of borel-measurable functions such that $f_{n}$ converges to $f$ almost everywhere on $A$ (relative to $\lambda$). Show that for each $M>0$ there is a $\lambda$-measurable $A_{M}\subseteq A$ such that $\lambda\left(A_{M}\right)>M$ and $f_{n}$ converges uniformly to $f$ on $A_{M}$.

My Thoughts: This is obviously similar to Egoroff's theorem and I have a feeling it could be used in the proof. I suppose from Egoroff's Theorem given $B\subseteq A$ such that $\lambda\left(B\right)<\infty$ for any $\varepsilon>0$ I can find a subset $C\subseteq B$ such that $\lambda\left(C\right)<\varepsilon$ and $f_{n}$ converges uniformly on $B\backslash C$ . I don't have a good idea on how to use this to prove the intended claim though.

2. Let $f_{n}:\left(\left[0,1\right],\mathcal{L}\right)\to\left(\mathbb{R},\mathcal{B}\right)$ be a sequence of Borel-measurable functions. Prove that there is a sequence $\left\{ c_{n}\right\} _{n=1}^{\infty}$ of positive real numbers such that $\frac{f_{n}}{c_{n}}\overset{n\to\infty}{\longrightarrow}0$ almost everywhere on $\left[0,1\right]$. [Hint: use Borel-Cantelli's Lemma] .

My Thoughts: As per the hint my line of thinking was to try and find a sequence of sets $\left\{ B_{n}\right\} _{n=1}^{\infty}$ such $\sum_{n=1}^{\infty}\lambda\left(B_{n}\right)<\infty$ and $$\limsup B_{n}=\left\{ x\in\left[0,1\right]\:|\:\lim\limits _{n\to\infty}\left(\frac{f_{n}\left(x\right)}{c_{n}}\right)\neq0\right\}$$ This given some choice of $\left\{ c_{n}\right\} _{n=1}^{\infty}$ which works. Unfortunately I have no idea what that choice might be. As for the $B_{n}$'s I imagine they would have to be something of the form $$\left\{ x\in\left[0,1\right]\,|\,\left|\frac{f_{n}\left(x\right)}{c_{n}}\right|>\frac{1}{n}\right\}$$ but I'm not really sure if this works.

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## 2 Answers

Proof of the second claim: To find an appropriate sequence $\{c_n\}$, fix $n$, and define $B_k=\{ x : \frac1k|f_n(x)|\geq\frac1n \}$. Then $B_k \searrow \emptyset$, and $\lambda(B_k)\rightarrow 0$. Therefore, there exists $k_n\in\mathbb N$ s.t. $\lambda(B_{k_n})<2^{-n}$.

Letting $c_n=k_n$, and $A_n=B_{k_n}$, we have (by Borel-Cantelli) $$\lambda (\limsup A_n)=0.$$ Letting $$L:=\{x : \lim\frac1{c_n}f_n(x)=0\},$$ we note that $L^\complement\subseteq \limsup A_n$. Hence, $\lambda(L^\complement)=0$, and $\frac1{c_n} f_n\overset{\text{a.e.}}{\longrightarrow} 0$.

Proof of the first claim: First we'll show that for all $M\in\mathbb{R}$ there is an $A_{M}\subseteq A$ such that $\lambda\left(A_{M}\right)>M$ , define:$\mathcal{A}=\left\{ \lambda\left(B\right)\;|\: B\subseteq A\quad\lambda\left(B\right)<\infty\right\}$ Now given $N\in\mathbb{N}$ we know that $\lambda\left(A\cap\left[-N,N\right]\right)\leq2N$ and thus $A\cap\left[-N,N\right]\in\mathcal{A}$ and we can see that:$$\infty=\lambda\left(A\right)=\lim_{N\in\mathbb{N}}\mathcal{\lambda}\left(A\cap\left[-N,N\right]\right)\leq\sup\left(\mathcal{A}\right)$$ Thus for all $M\in\mathbb{R}$ there is an $A_{M}\subseteq A$ such that $M<\lambda\left(A_{M}\right)<\infty$ . Now given $M>0$ and $\varepsilon>0$ we have a subset $A_{M+\varepsilon}\subseteq A$ such that $M+\varepsilon<\lambda\left(A_{M}\right)<\infty$ . From Egoroff's Theorem there is a $B\subseteq A_{M+\varepsilon}$ such that $\lambda\left(B\right)<\varepsilon$ and $f_{n}$ converges uniformly to $f$ on $A_{M+\varepsilon}\backslash B$ and additionally $$\lambda\left(A_{M}\backslash B\right)=\lambda\left(A_{M}\right)-\lambda\left(B\right)>\left(M+\varepsilon\right)-\varepsilon=M$$ Thus $A_{M}=A_{M+\varepsilon}\backslash B$ meets the requirements of the claim.

Still working on the latter claim.