# Characteristic Function Convergence problem

I have the following problem statement: Let $E \subset [0,1]$ be Lebesgue measurable, and for $x \in \mathbb{R}$, define $$f_n (x) := n \int_0^{1/n} \chi_{E} (x+t) dt$$ ($\chi$ denotes characteristic function). Show that $f_n$ is Lipschitz for every $n$, $f_n \to \chi_E (x)$ a.e, and $||f_n - \chi_E ||_{L^1} \to 0$.

I am not quite sure how to prove this is Lipschitz. It is clear that $0 \leq f_n \leq 1$ for all $n$. I can make a change of variable to see that $f_n (x) = \int_0^1 \chi_E (x+nt)$; proving $L^1$ convergence immediately gives that $f_n \to \chi_E$ a.e as well. I also see that with the above change of variable, it is intuitively clear at least that for sufficiently large $n$, $x+ nt \notin E$ so that the only term not "killed off" is $\chi_E (x)$. I am having trouble formally stating these observations though, any help is appreciated.

• Do you already know what is a Lebesgue point of a measurable set? Commented Nov 9, 2017 at 12:04

Lipschitz continuity is not hard to prove. Indeed, for every $h$ it holds $$|f(x+h)-f(x)| = n \left|\int_{x+h}^{x+h+1/n} \chi_E(t)\, dt - \int_x^{x+1/n} \chi_E(t)\, dt\right| \leq n \left|\int_x^{x+h} \chi_E + \int_{x+1/n}^{x+1/n+h} \chi_E\right| \leq 2n|h|,$$ hence $f_n$ is $2n$-Lipschitz continuous.

Since $$|f_n(x) - \chi_E(x)| \leq n \int_x^{x+1/n} |\chi_E(t) - \chi_E(x)|\, dt$$ you have that $f_n(x) \to \chi_E(x)$ at every Lebesgue point $x$ of $E$.

Finally, $$|f_n(x)| \leq 1, \qquad f_n(x) = 0 \ \forall x\not\in [-1,1],$$ so that, by the Dominated Convergence Theorem, $f_n \to \chi_E$ in $L^1$.

• are you sure of $[0,2]$? $f_n(-\frac 1{2n})$ is not necessarily zero.
– zwim
Commented Nov 9, 2017 at 12:32
• sorry, I have written $[0,2]$ instead of $[-1,1]$. Commented Nov 9, 2017 at 12:35
• I was not familiar with the theorem on Lebesgue points, thank you. I did not anticipate the use of that at all. Commented Nov 9, 2017 at 12:40