2

this is a homework problem and I can't figure it out.

  • Find a Borel measure $\mu$ on $\mathbb R$ such that for all $f \in L^+(\mathbb R)$, $\int_\mathbb R f \, d\mu = 0 \; \implies \; f = 0$

All of the notation is from Folland's Real Analysis book. If it helps, in the previous problems I showed under the same conditions:

  • $\{x \in X : f(x) = \infty \}$ is a null set.
  • $\{ x \in X : f(x) > 0 \}$ is $\sigma$-finite.

My guess is to use something simple like counting measure. But I really don't know where to start.

Michael Hardy
  • 1
Vergil
  • 21
  • 4
    You mentioned the counting measure. Why not start with checking whether that does the required things? – Daniel Fischer Nov 25 '14 at 17:23
  • Do you mean to go back to the definition of the integral and write it as a sup? I looked at http://math.stackexchange.com/questions/764076/integration-with-respect-to-counting-measure to see how to perform a computation with respect to counting measure, but R is uncountable. So should I use the Monotone Convergence Theorem? Does it even apply here? – Vergil Nov 25 '14 at 17:31
  • The integrability of the considered $f$ is part of the assumption [even if $L^+(\mathbb{R})$ allows infinite integral, you're only looking at functions with integral $0$], so you don't need to figure out what functions are integrable or wrestle with the definition of the integral. And, not that it's relevant for the question, but in general it is, the monotone convergence theorem holds for all positive measures. – Daniel Fischer Nov 25 '14 at 17:38
  • So then how else do I check whether or not the counting measure works? – Vergil Nov 25 '14 at 17:47
  • If $\nu$ is a positive measure, when is the integral of a non-negative measurable function $0$? – Daniel Fischer Nov 25 '14 at 17:49
  • When the measure of every positive value of the function is 0? – Vergil Nov 25 '14 at 17:55
  • That might be the right thing just weirdly expressed. Anyway, what are the null sets for the counting measure? – Daniel Fischer Nov 25 '14 at 18:20

1 Answers1

0

Let $\sigma_x$ denotes Dirac measure defined on the Borel class $B(R)$, i.e. for $A \in B(R)$, $\sigma_x(A)=1$ if $x \in A$ and $\sigma_x(A)=0$, otherwise. We set $\mu(A)=\sum_{x \in R}\sigma_x(A)$ for each $A \in R$. Then for $f \in L(R^{+})$ we get $$\int_Rf(y)d\mu(y)=\int_R f(y)d(\sum_{x \in R}\sigma_x)(y)= \sum_{x \in R}\int_R f(y)d(\sigma_x)(y)= \sum_{x \in R}f(x).$$ If $\int_R f(y)d\mu(y)=0$ then $\sum_{x \in R}f(x)=0$. Since $f \in L(R^{+})$ we have $f(x)\ge 0$ for each $x \in R$. Now it is obvious that $\sum_{x \in R}f(x)=0$ implies $f(x)=0$ for each $x \in R$.

Gogi Pantsulaia
  • 511
  • 2
  • 10