I am trying to prove the following Lemma, which seems intuitive, but I still have doubts:

**Lemma**

Given a Brownian motion $\{W_t,\mathcal F_t:0\le t \le1\}$, two bounded processes, $\mu$ and $\sigma$, with $\sigma$ continuous and $\sigma_0\neq 0$, such that the integral

$$ X_t=\int_0^t \sigma_t dW_t + \int_0^t \mu_t dt$$

exists, then $$\lim_{t\rightarrow 0}P\left(X_t >0\right)=\frac 1 2.$$

**Proof Attempt (wrong)**

Define $g_n := \sqrt{n} X_{1/n}$ and clearly it holds $P(X_{1/n}>0)=P(g_n>0)$, for all $n\in\mathbb N$. By the time change formula for stochastic integrals (Karatzas & Shreve, Th. 3.4.8) and regular integration by substitution we have:

$$ g_n = \int_0^1 \sigma_{s/n}dB_s + \frac 1 {\sqrt{n}}\int_0^1 \mu_{s/n}ds \quad\text{ a.s. }$$

where $B_s:=\sqrt{n} W_{s/n}$ is a Brownian motion. Furthermore, by bounded convergence, we have $g_n\stackrel{L^2}\longrightarrow \sigma_0 B_1$: *(as noted by @sinusx, this does not work as $B_1$ depends on $n$)*

$$ \mathbb E\left(g_n-\sigma_0 B_1\right)^2\le\mathbb E\left(\int_0^1 (\sigma_{s/n}-\sigma_0)^2ds+\frac 1 {n}\;\left(\int_0^1\mu_{s/n}ds\right)^2\right)\rightarrow 0$$

It remains to show that $P(g_n>0)=\mathbb E(1_{g_n>0})$ converges to $\mathbb E(1_{\sigma_0 B_1>0})$ $=$ $P(\sigma_0 B_1>0)=\frac 1 2$. By bounded convergence, it suffices to show almost sure convergence for $1_{g_n>0}$:

$L^2$ convergence implies almost sure convergence for $g_n$. As the step function is contiuous everywhere except at $0$, we conclude with $P(\lim_{n\rightarrow \infty}g_n=0)=P(B_1=0)=0$.

(BTW, this is the main difference between $g_n$ and $X_{1/n}$: $P(X_{1/\infty}=0)=1\neq P(\sigma_0 B_1=0)=0$.)

**Questions**

Is the Lemma true? Is the proof correct? Is there an easier proof or does it follow from some other result? Can you point me to similar results?

(Meta: Is question suitable for MO?)

**Edit: Simpler proof, suggested in @Sinusx's answer**

Define $f_t := \sigma_0\frac{W_t}{\sqrt{t}}$ and $g_t := \frac{X_t}{\sqrt{t}}$. It clearly holds

$$ g_t=f_t + \frac{1}{\sqrt{t} }\int _ 0 ^t (\sigma _s - \sigma _0 ) dW _s+ \frac{1}{\sqrt{t} } \int _0 ^t \mu _s ds, $$

implying $g_t - f_t\longrightarrow 0$ in $L^2$ by bounded convergence

$$ \mathbb E(g_t-f_t)^2 \le\mathbb E\left(\int_0^1 (\sigma_{st}-\sigma_0)^2ds+t\;(\sup_{s\le t}\mu_s)^2\right)\rightarrow 0, \text{ as } t\rightarrow 0 \tag{1}$$

and thus convergence in probability. Together with $f_t\sim \mathcal N(0,\sigma_0^2)$ (by the scaling property of the Brownian motion) we can apply this condition for convergence in distribution to show $g_t\stackrel{\text{(d)}}\longrightarrow \mathcal N(0,\sigma_0^2)$ as $t\rightarrow 0$. The lemma now follows from $P(X_t>0)=P(g_t>0)\rightarrow \frac{1}{2}$.

PS: However, I think (1) still needs a dominator for the $\sigma$ part.