# Construct a sequence of continuous functions $f_n: \mathbb{R} \rightarrow \mathbb{R}$such that $f_n$ converges to $f$ in $L_1 ({\lambda})$.

I am struggling to find a function that fits the criteria in this question.

Assume that $\Omega = \mathbb{R}$, $\mu = \lambda$ is the Lebesgue measure, and $\Sigma$ is the $\sigma$-algebra of Lebesgue measurable sets. Let $f=\chi_{[0,1]}: \mathbb{R} \rightarrow \mathbb{R}$ be the characteristic function of $[0,1]$. Construct a sequence of continuous functions $f_n:\mathbb{R} \rightarrow \mathbb{R}$ such that $f_n$ converges to $f$ in $L_1(\lambda)$. Justify your construction.

So it appears that for all $\epsilon \geq 0$, we need to find a sequence of continuous functions for which there exists an $N \in \mathbb{N}$ such that $|f_n - f| \leq \epsilon$.

I can't seem to think of something that is continuous!

• Approximate the "vertical" parts of the graph of $f$ by affine profiles with increasing slope. May 12, 2017 at 12:54
• but what would that look like in terms of a function? May 12, 2017 at 12:55

Hint: for $n\in\mathbb{N}$, $n\geq 2$, consider $$f_n(x) := \begin{cases} 0, & \text{if}\ x\leq 0\ \text{or}\ x\geq 1,\\ 1, & \text{if}\ 1/n\leq x \leq 1 - 1/n,\\ nx, & \text{if}\ 0 < x < 1/n,\\ n(1-x), & \text{if}\ 1-1/n < x < 1. \end{cases}$$