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](http://en.wikipedia.org/wiki/Proofs_of_convergence_of_random_variables#Convergence_in_probability_to_a_sequence_converging_in_distribution_implies_convergence_to_the_same_distribution) 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.



$\lim_{t\rightarrow 0}P\left(X_t >0\right)=\frac 1 2$ for continuous semimartingales?
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.


