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.

  • $\begingroup$ a lower bound is possible, not an upper bound. $\endgroup$ – Carlo Beenakker Dec 5 '19 at 22:32
  • $\begingroup$ @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. $\endgroup$ – Nate Eldredge Dec 6 '19 at 4:30
  • $\begingroup$ right, my mistake. $\endgroup$ – Carlo Beenakker Dec 6 '19 at 7:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.