I am a noob at measure theory so please forgive me for being naive, I just need help, here is what I have albeit this is a very very standard result:
Let $m$ denote Lebesgue measure on $\mathbb{R}$ and $E$ a measurable subset with FINITE measure.
Show that,
$\forall$ $\epsilon > 0$, $\exists$ open set $U$ containing $E$ s.t.
$m(U \backslash E)$ < $\epsilon$ (**)
What is the standard way to go about it because my intuition is that since $E$ is already measurable, then (**) becomes
$m(U \backslash E)$ $\leq$ $\epsilon$
then we assume further that $E$ has FINITE measure,
so $E$ is somewhat "small"er than if it has infinite measure, but how does this guarantee that the open set containing $E$ has a "smaller difference" to $E$ which in turn gives us $<$ and not $\leq$.
OR is there a subtlety on the fact that $m$ is Lebesgue measure and my problem only states $E$ is measurable, but when it states finite measure it says $m(E) < \infty$ which assumes $E$ is Lebesgue Measurable? so This thought can't be correct.
I am using Stein Shakarchi mainly although I have papa Rudin for reference.
Any helps is greatly appreciated, if theres ANYTHING I said incorrectly or am doing wrong, PLEASE LET ME KNOW!!