I have a really basic question that I believe may seem trivial to most of you, but I graduated in physics and am currently trying to get my head around some measure theory. Since the question involve just a part of the whole problem, I will paraphrase just this part.
...Let $E_m$ be a decreasing sequence of sets in a sigma algebra R, such that $ \lim_{n\rightarrow\infty}(E_m)=0 $
Thus, given an $\epsilon > 0 $ , there exists a number $n$ such that $ m(E_n)< \epsilon $.
Can someone tell me why the former implies the latter?
By definition of "$\lim_{n \to \infty} m(E_n) = 0$", we mean "given $\epsilon > 0$ there exists $N$ such that for all $n \ge N$ we have $|m(E_n)| < \epsilon$". This is the definition of the limit. Note in this case that since $(m(E_n))$ is a decreasing sequence (in $n$), you only need to find some $N$ such that $m(E_N) < \epsilon$, and then you know that for all $n \ge N$ we have $m(E_n) \le m(E_N) < \epsilon$, hence the result follows.
Note that if the limit in general were $L$ instead of $0$, then we replace "$|m(E_n)| = |m(E_n) - 0|$" by "$|m(E_n) - L|$". When the limit point is $0$, we can drop the modulus bars, since we know that $m(A) \ge 0$ for every set $A$, by definition of a measure $m$. However, we can't do this for a general limit $L$. Anyway, that's just some extra information, but hopefully it helps. :)