$L^2$ convergence implies uniform convergence? 


It is from Continuous Martingale and Brownian Motion. My question is why the highlight conclusion holds? I think in the proof of proving $L^2$ convergence is for fixed $t$. But now it conclude for all $t$ on $[0,a]$ and furthermore is uniformly convergence. Any help please.
 A: There is the following statement (see this answer for a proof):

Consider the space of stochastic processes with continuous sample paths such that the norm $$\|X\| := \left[\mathbb{E} \left( \sup_{t \leq a} |X_t|^2 \right)\right]^{\frac{1}{2}} \tag{1}$$ is finite for fixed $a>0$. Then this normed space is a complete space.

The estimate
$$\mathbb{E} \left( \sup_{t \leq a} |T_t^{\Delta_n}-T_t^{\Delta_m}|^2 \right) \leq 4 \mathbb{E}((T_a^{\Delta_n}-T_a^{\Delta_m})^2) \xrightarrow[]{n,m \to \infty} 0$$
shows that
$$X_n(t,\omega) := T_t^{\Delta_n}(\omega)$$
is a Cauchy sequence with respect to the norm defined in $(1)$; hence convergent, i.e. there exists a stochastic process $X$ with continuous sample paths such that
$$\mathbb{E} \left( \sup_{t \leq a} |X_n(t)-X(t)|^2 \right) \to 0.$$
This, however, means that $\sup_{t \leq a} |X_n(t)-X(t)|$ converges in $L^2$ to $0$, and therefore there exista a subsequence which converges almost everywhere to $0$:
$$\sup_{t \leq a} |X_{n_k}(t)-X(t)|^2 \to 0 \qquad \text{a.s.}$$
