So I have a few questions on convergence in measure (Lebesgue measure)
So we know that convergence on measure is induced by the norm $$d(f,g)=\int_E \frac{|f-g|}{1+|f-g|}$$ on finite measure $E$
Thus, we know that if $f_n$'s every sub-sequence has a convergent sub-sequence, then the sequence converges since we are on a topological space, correct? Are there any other neat facts about convergence in measure that can be deduced from the fact the we have a normed vector space?
Also on finite measure it can be shown that $f_n \to f$ in measure $\iff$ every sub-sequence has a further sub-sequence that converges a.e. Does this simply follow from the preceding observation? As pointwise implies in measrue on finite measure space?
Is this the standard proof ?:
Assume that $f_n \not \to f$, then $\lambda(|f_{n_k}-f|\geq \delta)\geq \epsilon$ for some $\delta,\epsilon$ and all $k$. But now we can pick a subsequence of $f_{n_k}$ that converges a.e and by egorof we know it converges in measure. Thus we contradict $\lambda(|f_{n_k}-f|\geq \delta)\geq \epsilon$
Once you know convergence in measure is equivalent to convergence in that metric, every familiar fact about metric spaces is fair game.
However, do note that $d$ is a metric, not a norm. And $\|f\|:=d(f,0)$ is not a norm because scalars don't factor out properly. So, convergence in measure is metrizable, not necessarily normable.
To see that there is no norm on $L^1$ such that convergence in that norm is equivalent to convergence in measure, we can argue as follows:
Consider Lebesgue measure on $[0,1]$. By a theorem of Nikodym there is no continuous linear functional $\phi: (L^1([0,1]),d)\to \mathbb{R}$ except for the identically $0$ one. For a (sketch of) a proof, see Bogachev's Measure Theory, volume 1, exercise 4.7.61 on page 306. By the Hahn-Banach theorem, the metric topology induced by $d$ cannot be induced by a norm, and thus neither can convergence in measure.