$\lim_{t\rightarrow 0}P\left(X_t >0\right)=\frac 1 2$ for continuous semimartingales? 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.
 A: The proof is not correct, as without additional integrability condition you will not be able to conclude that $g_n \in L^2$ for $n$ large enough, and therefore the $L^2$ convergence argument fails. As for a concrete counterexample, something like $\sigma_t = e^{e^{W_t}}$ (and $\mu_t =0$) should do the trick.
A: I think that the lemma is true if you add some additional continuity assumptions on $\sigma$. In the proof you are using the right scaling, however you may want to replace convergence in $L^2$ with convergence in distribution, as $g _ \infty$ is not defined and does not exist. Note that $\frac{1}{\sqrt{t}} W _{t} \overset{(d)}{=}W_1$, so $\frac{1}{\sqrt{t}} W _{t} $ converges in distribution to $N(0,1)$, $t \to 0$, but  $ \limsup \frac{1}{\sqrt{t}} W _{t} = \infty$,  $ \liminf \frac{1}{\sqrt{t}} W _{t} = - \infty$ by the law of iterated logarithm.
For example, let in addition to your assumptions $\sigma$ be mean square continuous, that is, $E |\sigma _s| ^2 < \infty$ for all $s$ and and $E|\sigma _t - \sigma _0|^2 \to 0$, $t \to 0$.
We have
$$
\frac{X _t}{\sqrt{t} } =\sigma _0 \frac{W _t}{\sqrt{t} } + \frac{1}{\sqrt{t} }\int _ 0 ^t (\sigma _s - \sigma _0 ) dW _s+ \frac{1}{\sqrt{t} } \int _0 ^t \mu _s ds,
$$
 You can prove then that the second and third summands on the right hand side go to $0$ in $L^2$ as $t \to 0$, but the first does not and you can get the statement of the lemma.
