Property of a finite positive measure on $\mathbb{R}$

Question

Let $m$ be a positive finite measure on $\mathbb{R}$ and $E\subset \mathbb{R}$ such that $$m(E)=0.$$ Can we conclude that Lebesgue measure of $E$ is also zero?

Can we conclude that if Lebesgue measure of $E$ is zero then $m(E)=0?$

Answer 1

The property you mention is called absolute continuity of measures and it is not true in general that a positive finite measure must be absolutely continuous w.r.t. the lebesgue measure or viceversa. The usual example is a singular measure like the dirac measure $$m(E)=\begin{cases} 1& \text{if $0\in E$} \\ 0 & \text{if $0\notin E$} \end{cases}$$ It's easy to see that this is a finite positive measure and that $\{0\}$ is set of lebesgue measure 0 but dirac measure 1 and $(1,2)$ is a set of lebesgue measure 1 but dirac measure 0.

However you can always decompose a measure into an absolutely continuous part and a singual part https://en.m.wikipedia.org/wiki/Lebesgue%27s_decomposition_theorem#:~:text=Hence%20(singular%20continuous%20measures%20aside,of%20a%20singular%20continuous%20measure.

Answer 2

The answer is NO to both. Let $m(A)=\lambda (A \cap (0,1))$ where $\lambda $ is the Lebesgue measure. Let $E=(-1, 0)$. Then $m(E)=0$ but $\lambda (E)=1$.

Let $m$ be the Dirac measure at $0$: $m(A)=1$ if $ 0 \in A$ and $0$ otherwise. Then $m((0,1))=0$ but $\lambda ((0,1))=1$.