Let \(E,F\subset\mathbb R\) be measurable sets of finite measure.
Show that \(\chi_E*\chi_F\) is continuous.
Show that \[ n\bigl(\chi_E*\chi_{[0,1/n]}\bigr)\to\chi_E \] pointwise almost everywhere.
Solution
1. The convolution is continuous.
Since \(E\) and \(F\) have finite measure, \[ \chi_E,\chi_F\in L^2(\mathbb R). \] For \(h\in\mathbb R\), \[ \begin{aligned} | (\chi_E*\chi_F)(x+h)-(\chi_E*\chi_F)(x) | &\le \|\chi_E\|_2\,\|\chi_F(\cdot+h)-\chi_F\|_2. \end{aligned} \] The right-hand side is independent of \(x\) and tends to \(0\) as \(h\to0\) by continuity of translations in \(L^2\). Hence \(\chi_E*\chi_F\) is uniformly continuous, in particular continuous.
2. The normalized interval convolutions converge almost everywhere to \(\chi_E\).
For every \(x\), \[ \begin{aligned} n(\chi_E*\chi_{[0,1/n]})(x) &=n\int_{\mathbb R}\chi_E(y)\chi_{[0,1/n]}(x-y)\,dy\\ &=n\int_{x-1/n}^{x}\chi_E(y)\,dy. \end{aligned} \] Thus this is the average of \(\chi_E\) over the interval \([x-1/n,x]\). By the one-sided Lebesgue differentiation theorem, for almost every \(x\), \[ n\int_{x-1/n}^{x}\chi_E(y)\,dy\longrightarrow\chi_E(x). \] This proves the claim.