Let \(E\subset\mathbb R\) be measurable.
If \(|E|<\infty\), prove that \[ F(x)=\int_{\mathbb R}\chi_E(y)\chi_E(y-x)\,dy \] is continuous.
If \(|E|>0\), possibly infinite, prove that \(E-E\) contains an interval \((-\varepsilon,\varepsilon)\) for some \(\varepsilon>0\).
Solution
1. Continuity when \(E\) has finite measure.
Since \(|E|<\infty\), one has \(\chi_E\in L^2(\mathbb R)\). For \(h\in\mathbb R\), \[ \begin{aligned} |F(x+h)-F(x)| &=\left|\int \chi_E(y)\bigl(\chi_E(y-x-h)-\chi_E(y-x)\bigr)\,dy\right|\\ &\le \|\chi_E\|_2\,\|\chi_E(\cdot-h)-\chi_E\|_2. \end{aligned} \] The inequality is Cauchy--Schwarz in $L^2$. The right-hand side is independent of \(x\) and tends to \(0\) as \(h\to0\) by continuity of translation in $L^p$. Thus \(F\) is uniformly continuous.
2. Positive-measure sets have a neighborhood of zero in the difference set.
Assume \(|E|>0\). Because Lebesgue measure is sigma-finite, there exists \(R>0\) such that \[ A=E\cap[-R,R] \] has finite positive measure. Apply part (a) to \(A\) and define \[ F_A(x)=\int\chi_A(y)\chi_A(y-x)\,dy. \] Then \[ F_A(0)=|A|>0. \] By continuity, there exists \(\varepsilon>0\) such that \[ F_A(x)>0 \qquad(|x|<\varepsilon). \] If \(F_A(x)>0\), then the set of \(y\) with \[ y\in A, \qquad y-x\in A \] has positive measure and is therefore nonempty. For such a \(y\), \[ x=y-(y-x)\in A-A\subset E-E. \] Hence \[ (-\varepsilon,\varepsilon)\subset E-E. \]
3. Q.E.D.
Step 1 proves part (a), and step 2 proves part (b).