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.