$L^1$ convergence of a sequence of stochastic integrals and convergence of their quadratic variations On a filtered probability space $(\Omega, \mathcal F, \mathcal F_t, \mathbb P)$ containing a Brownian motion $W_t$. Let $\sigma^n_t >0$ be a sequence of square intergable adapted processes and consider:
 $$Z_t^n=\int_0^t \sigma^n_s d W_s $$
Assume $Z_t^n$ to be such that:
$$ \langle Z_t^n \rangle = \int_0^t (\sigma_s^n)^2 ds \rightarrow_{L^1(\Omega)} \alpha_t $$
for all $t$ and for some predictable process $\alpha_t$. 
Is it true that, for all $t$:
$$  Z_t^n \rightarrow_{ L^1(\Omega) } Z_t $$
where $\langle Z_t \rangle = \alpha_t$?. In other words does the $L^1$ convergence of the quadratic variations imply convergence of the stochastic integrals, and if so is the limit of the quadratic variations the quadratic variation of the limit?
Failing this, does the second statement hold true if we have established the existence of an $L^1$ limit for $Z_t^n$ already? Also I am happy to replace $L^1$ with $L^2$ (or $L^p$).
This seems very much like a Levy theorem/martingale characterization of quadratic variation sort of problem, but I cannot quite handle it.  
 A: This is the best answer I came up with. 
Assume \sigma_t^n \rightarrow \sigma_t $(\sigma_t^n)^2 \rightarrow \sigma^2_t$  in $L^1(\Omega)$ for all $t$, and let $Z_t= \int^t_0 \sigma_s dW_s$. We have, using the monotonicity of $L^p$ norms, the Ito isometry and $(a-b)^2 \leq |a^2-b^2|$: 
$$||Z_t^n- Z_t ||_1 = \Big | \Big|\int _0^t (\sigma^n_s - \sigma_s)dW_s \Big | \Big|_1 \leq \mathbb E \left[\int _0^t (\sigma^n_s -\sigma_s)^2 d_s \right]^{1/2} \leq \mathbb E \left[\int _0^t |((\sigma^n_s)^2-\sigma^2_s)| d_s \right]^{1/2}=\left( \int _0^t \mathbb E[ |((\sigma^n_s)^2-\sigma^2_s)|] d_s \right)^{1/2} $$
Maybe I need some mild extra assumptions to apply Application of Fubini's theorem in the last line is justified if $\mathbb E \left[ |((\sigma^n_s)^2-\sigma^2_s)| \right]$ is Riemann integrable/continuous (always true for expectations of continuous stochastic processes? No problems of uniform integrability in $\Omega$?).  Taking the limit inside the integral is ok I believe  We take the limits in the above, which in the right hand side can be further interchanged with integration because the integrands are continuous and converge pointwise in the $s$ variable by assumption. The $L^1(\Omega)$ convergence of $(\sigma_t^n)^2$ to $\sigma_t^2$ yields that $$Z_t^n \rightarrow_{L^1(\Omega)} Z_t$$
EDIT: Edited to answer TheBridge's comments below
