Integration over a Translated Set is Uniformly Continuous?

Question

I'm looking for help with the following problem (working with Lebesgue measure here):

Let $ψ∈L^\infty(R)$ and let $B \subseteq \mathbb R $ be measurable such that $m(B) < \infty$. If $B+x =\{b+x|b∈B\}$ and if $$F(x) =\int_{B+x}ψ(v)dv$$ then prove that $F$ is uniformly continuous on $\mathbb R$.

I want to argue that if $|x - y| < \delta$ then $|F(x) - F(y)| < \epsilon$. Intuitively, it feels as though $|F(x) - F(y)| < \delta \cdot \| \psi\|_{L^{\infty}}$

The thought is that $F(x)$ and $F(y)$ should only differ on a set of measure $<\delta$. (More accurately, it feels like I should be able to choose a $\delta$ small enough that this is the case). However I haven't been able to show anything like this and I can't seem to work around it to find a proof either.

Answer 1

Consider that $\displaystyle F(x)=\int_{\Bbb{R}}\! \psi(v)\mathbf{1}_{B+x}(v)\, \mathrm{d}v$ (where $\mathbf{1}_X$ is the characteristic function of $X\subset\Bbb{R}$).

This can be rewritten as $\displaystyle F(x)=\int_{\Bbb{R}}\! \psi(v)\mathbf{1}_{-B}(x-v)\, \mathrm{d}v=(\psi*\mathbf{1}_{-B})(x)$.

We may assume that $m(B)>0$ (since: if $B$ is a null set then $F=0$ ). Now your problem has been addressed in this very similar question.

Answer 2

There's a solution rather similar to jp boucheron's but simpler. Write $$F(x) = \int_{\mathbb{R}} \psi(v) \textbf{1}_{B} (v-x) \, dv = \int_{\mathbb{R}} \psi(v) K(v-x) \, dv$$ where $K = \textbf{1}_{B} \in L^1(\mathbb{R})$ since $B$ has finite measure. Here I've used the trivial to check fact that $\textbf{1}_{B + x} (v) = \textbf{1}_{B} (v-x)$.

Then \begin{align*} |F(x) -F(y)| & = \left| \int_{\mathbb{R}} \psi(v) K(v-x) \, dv - \int_{\mathbb{R}} \psi(v) K(v-y) \, dv\right| \\ & = \left| \int_{\mathbb{R}} \psi(v)(K(v-x) - K(v-y)) \, dv\right| \\ & \leq \|\psi\|_{L^\infty} \int_{\mathbb{R}} |K(v-x) - K(v-y)| \, dv \\ & = \|\psi\|_{L^\infty} \int_{\mathbb{R}} |K(u) - K(u+ x - y)| \, dv\end{align*} where in the last line I've made the substitution $u = v - x$. But it's a standard result (proved via $L^q$ density of step functions, or density of continuous, compactly supported functions, whichever you prefer) that $$\lim_{s \rightarrow 0} \int_{\mathbb{R}} |K(u) - K(u+s)|^q \, dv = 0$$ for any $K \in L^q(\mathbb{R})$ for $1 \leq q < \infty$.

The point here is: doing this problem assuming $\psi \in L^\infty(\mathbb{R})$ makes convolution unnecessary. On the other hand to do this assuming $\psi \in L^p(\mathbb{R})$ for $1 \leq p < \infty$ would require one to do exactly what jp boucheron does and then use Young's convolution inequality.

Answer 3

Hints: $\psi$ is a uniform limit of simple functions. When $\psi=\chi_E$ we get $F(x)=\mu ((B+x)\cap E)$ and $|F(x)-F(y)|\leq 2\mu (B)$ when $B$ is an interval (where $\mu$ is Lebesgue measure). Now use the basic approximation of Measure Theory which says that you can approximate $B$ by a finite union of intervals. [Ref. Measure Theory by Halmos].