We know that $L_1$ convergence does not imply convergence a.e. in general. However, consider the following set
$$S=\{f:[0,1] \rightarrow [0,1]\mid f(x)\geq f(y) \quad \forall x\geq y\}.$$ Take any sequence $\{f_n\}_1^\infty\subset S$ that is converging to $f$ with respect to $L_1$ norm. Is it possible that convergence in $L_1$ implies pointwise a.e. convergence in this case? I just couldn't find an example to show this is not true.
Does $L_1$ Convergence imply almost everywhere convergence for the Set of all increasing functions on $[0,1]$ to $[0,1]$
functional-analysismeasure-theoryreal-analysis
Best Answer
Yes.
Lemma. Suppose $f\in S$, $x\in(0,1)$, $f$ is continuous at $x$ and $\epsilon>0$. There exists $\delta>0$ such that if $g\in S$ and $|g(x)-f(x)|>\epsilon$ then $||f-g||_1\ge\epsilon\delta/2$.
Proof: Suppose first $g(x)-f(x)>\epsilon$. Choose $\delta>0$ so that if $I=(x,x+\delta)$ then $I\subset(0,1)$ and $$|f(t)-f(x)|<\epsilon/2\quad(t\in I).$$If $t\in I$ then $$g(t)-f(t)\ge g(x)-(f(x)+\epsilon/2)>\epsilon/2,$$so $$||g-f||_1\ge\epsilon\delta/2.$$
If on the other hand $g(x)-f(x)<-\epsilon$ a similar argument works with $I=(x-\delta,x)$.
Corollary. If $f,f_n\in S$ and $||f_n-f||_1\to0$ then $f_n\to f$ almost everywhere on $[0,1]$.
Because a monotone function is continuous almost everywhere.
Exercise. Explain the relevance of the example $$f_n(x)=\begin{cases} -1+nx,&(0\le x\le 1/n), \\0,&(1/n\le x\le 1).\end{cases}$$