# Problem

Suppose $f_n \to 0$ in measure and $f_n(x) \geq 0$ for each $x \in X$. Prove that $g(x) := \underline{\lim} f_n(x)$ we have $g(x)=0$ almost everywhere.

Give an example of a sequence of Lebesgue measurable non-negative functions $f_n:[0,1] \to [0,\infty)$ such that $f_n \to 0$ in lebesgue measure, but $g(x) := \overline{\lim}f_n(x)$ we have $g(x)=1$

# Attempt

Since $f_n \to 0$ in measure we can conclude there is a sub-sequence $n_k$ where $f_{n_k} \to 0$ almost everywhere. I feel like this gives the result but there is more to say?

I have no idea on the example

Thanks

In the first part you can apply (or repeat) the Borel-Cantelli lemma.

Since $f_n\to0$ in measure, there is an increasing sequence $n_k$ such that $\mu\big(f_{n_k}\ge\tfrac1{2^k}\big)<\tfrac1{2^k}$. We will show that $f_{n_k}\to0$ a.e.

Let $A_k=\{x\colon f_{n_k}(x)\ge\tfrac1{2^k}\big\}$; so $\mu(A_k)<\tfrac1{2^k}$; for every $m$ we have $$\Big\{ x\colon \varlimsup_{k\to\infty} f_{n_k}(x)\ge \tfrac1{2^k}\Big\} \subset\bigcup_{\ell=m}^\infty A_\ell$$ $$\mu\Big(\big\{ x\colon \varlimsup_{k\to\infty} f_{n_k}(x)\ge \tfrac1{2^k}\big\} \Big) \le \sum_{\ell=m}^\infty \mu(A_\ell) < \tfrac2{2^m}.$$ From $m\to\infty$ we get $$\mu\Big(\big\{ x\colon \varlimsup_{k\to\infty} f_{n_k}(x)\ge \tfrac1{2^k}\big\} \Big) =0,$$ therefore $$\mu\Big(\big\{ x\colon f_{n_k}(x)\not\to0 \big\} \Big) \le \sum_{k=1}^\infty \mu\Big(\big\{ x\colon \varlimsup f_{n\to\infty} f_n(x)\ge \tfrac1{2^k} \big\}\Big)= 0.$$

For the second part, For every $0\le k<m$, let $$f_{m,k}(x) = \begin{cases} 1 & \tfrac{k-1}{m} \le x \le \tfrac{k}{m}\\ 0 & \text{otherwise}\end{cases}$$ and order them in any order. This sequence tends to $0$ in measure, but for every $x\in[0,1]$, $f_{k,m}(x)=1$ infinitely many times.

• I'm not sure I completely understand the last couple lines – Aaron Zolotor Jan 8 '18 at 16:47
• For every $x$, there are infinitely many pairs $m,k$ s.t. $f_{k,m}(x)=1$, so the limsup is $1$. – user141614 Jan 8 '18 at 20:46
• Sorry, I didn't mean the example. The last couple lines of your proof. – Aaron Zolotor Jan 8 '18 at 23:16
• Edited: There was a typo, a liminf instead of limsup. $(\lim f\ne0)=\bigcup(\varlimsup f\ge1/2^k)$. – user141614 Jan 9 '18 at 6:00

Your proof for the first part is fine. For the second part, let $X=[0,1]$. Consider the "dansing sequence" that is defined as: \begin{align} f_1(x)&=\mathbf{1}_{[0,\frac{1}{2}]}\\ f_2(x)&=\mathbf{1}_{(\frac{1}{2},1]}\\ f_3(x)&=\mathbf{1}_{[0,\frac{1}{4}]}\\ f_4(x)&=\mathbf{1}_{(\frac{1}{4},\frac{1}{2}]}\\ \vdots \end{align} You know how this goes. You can come up yourself to the general terms if you would like. This function converges nowhere pointwise, but it converges to $0$ in measure. Since each point gets "visited" infinitely many times by the indicator function, you can see that for each particular $x\in X$ one can construct a subsequence $f_{n_k}$ such that $f_{n_k}(x)=1$ for all $k$. That means that for each $x\in X$ we have $\limsup_{n}f_n(x)=1$. All this can be made formal ofcourse, but this is the idea.