Consider $dS_t = \mu S_t dt + \sigma S_t dW_t$ with initial $S_0 > 0$. We may obtain that $S_t = S_0 \exp\left[(\mu - \frac{1}{2}\sigma^2) t +\sigma W_t\right]$. Hence we may consider average value of $S_T$. $A(T) = \displaystyle \frac{1}{T}\int_0^T S_0 e^{(\mu - \frac{1}{2}\sigma^2)t + \sigma W_t} dt$.
We want to determine $\lim_{T \to 0} A(T)$.
The main problem that we can't manage substitutions like $t = Tu$, because it's SDE. We should be careful about $W_t$. I've tried to consider $\xi \stackrel{d}{=} e^{\sigma W_t}$ and obtain $\mathbb{E} \displaystyle \frac{1}{T}\int_0^T \xi e^{(\mu - \frac{1}{2}\sigma^2)t}dt$. But that gives me limit in $L_1$. Not actually we interested in.
Any ideas? Maybe I miss something?
The fundamental theorem of calculus shows that
$$\lim_{T \to 0} \frac{1}{T} \int_0^T f(t) \, dt = f(0)$$
for any continuous function $f$. Applying this result for $f(t):=S_0(\omega) \exp\left(t \left[\mu-\frac{\sigma^2}{2} \right] + \sigma W_t(\omega) \right)$ with $\omega$ fixed, we find that
$$\lim_{T \to 0}\frac{1}{T} \int_0^T S_0(\omega) \exp\left(t \left[\mu-\frac{\sigma^2}{2} \right] + \sigma W_t(\omega) \right) \, dt = S_0(\omega).$$