Let $\mu$ be the outer measure, then...

Question

If $A\subset\mathbb{R}$ and $\mu(A)>0$ then exits $B\subset{A}$ bounded such that $\mu(B)>0$. I know the proof with $B$ a nonmensurable set, but can not B be simply subset of A without needing to be a nonmeasurable set?...and if it is true how do you prove it?

Answer 1

$A$ is the union of the sets $A \cap (-n,n)$, $n=1,2,...$. If each set on the right has outer measure 0 then A would have outer measure 0 because outer measures are sub-additive. Hence there exists n such that $B \equiv A \cap (-n,n)$ has positive measure.

Answer 2

If I am interpreting the question correctly, you are asking the following: for arbitrary measurable $A \subset \mathbb R$ with $\mu(A) > 0$, can we find a measurable $B \subset A$ with $\mu(B) > 0$?

The answer is yes. Trivially, we could take $B = A$.

If instead, you want $0 < \mu(B) < \mu(A)$, this will depend on the measure. If the measure has atoms, then no, we can't necessarily this. If the measure is atomless (like the Lebesgue measure), then we can accomplish this. Define $$F(x) = \mu(A \cap (-\infty, x]), \,\,\,\, x \in \mathbb R.$$ This function is continuous with $$\lim_{x \to -\infty} F(x) = 0, \,\,\,\,\,\,\, \lim_{x\to \infty} F(x) = \mu(A)$$ so by the intermediate value theorem, for any $c \in (0,\mu(A))$, we can find $x$ such that $\mu(A \cap (-\infty, x]) = F(x) = c$.

Then put $B = A \cap (-\infty, x]$.