# How to prove that there exists $M\lt\infty$ such that $\lvert \mu\rvert (E)\le M\lvert E\rvert$ for any Borel $E\subset [0,1]$.

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).

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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.