Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I need help in solving my homework problems. I would appreciate any effort!

Let $\mu$ be a signed Borel measure on $[0,1]$ and let $\lvert \cdot \rvert$ be a Lebesgue measure. Suppose that for any $\lvert\cdot \rvert$–summable function $f\colon [0,1] \to \mathbb{R}$ we have $\int_{[0,1]} f \,\text{d}\mu\ \lt \infty.$ Prove that there is a finite number $M$ such that $\lvert \mu\rvert (E)\le M \lvert E\rvert$ for any Borel subset $E$ of $[0,1]$.

I believe the fact $\lvert \nu\rvert(E)=\sup\{\int_E f d\nu\, \colon \lvert f\rvert\le 1\}$ may be useful. However, I do not know in which way (if at all).

share|cite|improve this question

Sketch of proof: defining $$F_n:=\{f\in L^1(\lambda),\int_{[0,1]}fd\mu\leqslant n\},$$ show that $F_n$ is closed for the $L^1(\lambda)$ norm and that $\bigcup_nF_n=L^1(\lambda)$. This gives that the embedding $\iota\colon L^1(\lambda)\to L^1(\mu)$ is continuous.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.