# Estimate Integral w.r.t Dirac measure by Integral w.r.t Lebesgue measure

Given a bounded, nonnegative and measurable function $$f$$ and denote the Euclidean ball with center $$z$$ and radius $$r$$ by $$B_r(z)$$.

Is it somehow possible to estimate the Integral w.r.t. the Dirac Lebesgue measure by the Integral w.r.t the Lebesgue measure?

I.e. is something like $$\int_{B_r(z)} f(x) \delta_z (dx) \leq \int_{B_r(z)} f(x) dx$$ true (maybe with some positive constant on the r.h.s)?

I believe its false, because the l.h.s is simply $$f(z)$$ (which we can assume to be positive and not zero) while the r.h.s might be zero.

• a lower bound is possible, not an upper bound. – Carlo Beenakker Dec 5 '19 at 22:32
• @CarloBeenakker: I'm not sure what kind of lower bound you have in mind, but taking $f = a \cdot 1_{\{z\}^C}$ seems to rule out such a thing. – Nate Eldredge Dec 6 '19 at 4:30
• right, my mistake. – Carlo Beenakker Dec 6 '19 at 7:15