A problem occur to me when I was learning real analysis:
Can we construct a measure $\mu$ on $\mathbb{R}$ such that for all $E\subset\mathbb{R}$, $E$ is $\mu$-measurable?
I know that for Lebesgue measure, this is not true. There are many non-measureable sets (with respect to Lebesgue measure) on $\mathbb{R}$. However, I wonder whether we can construct a measure $\mu$ such that there are no $\mu$-nonmeasurable sets.
I was thinking about adding non-measurable sets directly into $m$, i.e., define $\mu(E)=m(E)$ if $E$ is Lebesgue measurable and $\mu(E)=0$ if $E$ is not Lebesgue measurable. However, I was not sure that $\mu$ is still a measure after such modification.