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In this link math.stackexchange.com/questions/522613, it is stated that the sequence of positive continuous functions $(g_n)$ converges towards $0$ in measure and that one can extract a subsequence that converges a.e.

The functions are defined on $I=[0,1]$, and each $g_n$ is null eveywhere except on each $\left[\dfrac{k}{n},\dfrac{k}{n}+2^{-n}\right] (k=0,1,\cdots,n-1)$

However, is it mandatory to extract a subsequence ? Can't we state that $(g_n(x))$ converges a.e. towards $0$ ? Here's my idea, can anyone tell me if it is correct, or, where it is wrong ?

Let $m$ be the Lebesgue measure and $N$ be the set of all $x\in[0, 1]$ where $(g_n(x))$ doesn't converge to $0$. Let $S_n$ and $R_n$ be these sets : $$S_n=\bigcup_{k=0}^{n-1}\left[\dfrac{k}{n},\dfrac{k}{n}+2^{-n}\right]\qquad\text{and}\qquad R_n=\bigcup_{p>n}S_p$$

It appears that $$m(S_n)=\frac{n}{2^n}\qquad\text{and}\qquad m(R_n)\leq\sum_{p>n}m(S_p)=\sum_{p>n}\frac{p}{2^p}$$

The series which general term is $n2^{-n}$ converges. Hence, for any $\varepsilon>0$, there exists $n_0\in\mathbb{N}$ such as $m(R_{n_0})<\varepsilon$

Also, by definition of $R_n$ and $S_p$, $$\forall x\in[0, 1]\setminus R_{n_0}\quad,\quad\forall p>n_0\quad,\quad g_p(x)=0$$ which leads to $$\forall x\in[0, 1]\setminus R_{n_0}\quad,\quad \lim_{n\rightarrow+\infty}g_n(x)=0$$

Thus $N\subset R_{n_0}$ and this leads to $$\forall\varepsilon>0\quad,\quad m(N)<\varepsilon$$ As this is true for any $\varepsilon>0$ this gives $m(N)=0$, which means that $(g_n(x))$ converges towards $0$ a.e.

hdci
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1 Answers1

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Indeed your proof is correct. Another way to see this is by considering the set
$$S^*=\limsup_nS_n=\bigcap_{n\geq1}\bigcup_{k\geq n}S_k$$ $S^*$ is the set of $x\in I=[0,1]$ that belong to infinitely many $S_n$'s. Thus, for $x\in I\setminus S^*$, there is $n_x\in\mathbb{N}$ such that $x\in I\setminus S_n$ for all $n>n_x$ and so, by definition $g_n(x)=0$ for all $n>n_x$.

$$m(S^*)=\lim_nm\Big(\bigcup_{k\geq n}S_k\Big)\leq \lim_n\sum_{k\geq n}m(S_k)=\lim_{n\rightarrow\infty}\sum_{k\geq n}k2^{-k}=0$$ This shows that $g_n\xrightarrow{n\rightarrow\infty}0$ $m$-a.s. in $I$.

Mittens
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