Suppose \(F \subseteq \RR\) is closed with \(m(F^c) < \infty\) and let \(\delta(x) \da d(x, F)\) and \[ I_F(x) \da \int_\RR { \delta(y) \over \abs{x-y}^2 } \dy .\]
Show that \(\delta\) is continuous.
Show that if \(x\in F^c\) then \(I_F(x) = \infty\).
Show that \(I_F(x) < \infty\) for almost every \(x\in F\).
Solution
1. The function \(\delta(x)=d(x,F)\) is \(1\)-Lipschitz, hence continuous.
For \(x,x'\in\RR\) and every \(z\in F\), the triangle inequality gives \[ d(x,z)\le |x-x'|+d(x',z). \] Taking the infimum over \(z\in F\) yields \[ \delta(x)\le |x-x'|+\delta(x'). \] Interchanging \(x\) and \(x'\) gives \[ |\delta(x)-\delta(x')|\le |x-x'|. \]
2. If \(x\in F^c\), then \(I_F(x)=\infty\).
Because \(F\) is closed, \(F^c\) is open, so \[ r\definedas\delta(x)=d(x,F)>0. \] By step 1, whenever \(|y-x|<r/2\), \[ \delta(y)\ge \delta(x)-|x-y|>r/2. \] Therefore \[ I_F(x) \ge {r\over2}\int_{x-r/2}^{x+r/2}{dy\over|x-y|^2} =\infty. \]
3. For every \(y\in F^c\),\[ \int_F {dx\over|x-y|^2}\le {2\over\delta(y)}. \]
If \(y\in F^c\), then \(\delta(y)>0\) because \(F\) is closed. By the definition of distance, \[ |x-y|\ge\delta(y) \qquad\text{for every }x\in F. \] Hence, after translating by \(y\), \[ \int_F {dx\over|x-y|^2} \le \int_{|t|\ge\delta(y)}{dt\over t^2} ={2\over\delta(y)}. \]
4. The function \(I_F\) is integrable over \(F\) and\[ \int_F I_F(x)\,dx\le 2m(F^c). \]
Define the value on the diagonal \(x=y\) to be \(0\); this changes no integral. The integrand is then nonnegative and measurable, so Tonelli’s theorem gives \[ \begin{aligned} \int_F I_F(x)\,dx &=\int_F\int_\RR {\delta(y)\over|x-y|^2}\,dy\,dx\\ &=\int_\RR\left(\int_F{\delta(y)\over|x-y|^2}\,dx\right)dy. \end{aligned} \] For \(y\in F\), the inner integrand is \(0\) almost everywhere in \(x\), so its integral is \(0\). For \(y\in F^c\), step 3 gives \[ \int_F{\delta(y)\over|x-y|^2}\,dx =\delta(y)\int_F{dx\over|x-y|^2}\le2. \] Thus \[ \int_F I_F(x)\,dx \le\int_{F^c}2\,dy =2m(F^c)<\infty. \]
5. \(I_F(x)<\infty\) for almost every \(x\in F\).
By step 4, the nonnegative measurable function \(I_F|_F\) has finite integral. Therefore it is finite almost everywhere on \(F\).