Let $(\Omega, \mathcal{F}, P)$ be a probability space, on which
$\mathcal{F}\_t$ is filtration satisfying general conditions.
$W\_t$ is
a standard Brownian motion.
Let $Y\_t$ be a martingale given by
$$Y\_t = \int\_0^t \sigma\_r d W\_r$$
where $\sigma\_t$ is a bounded $\mathcal{F}\_t$ measurable process.


The question is,
assume $\sigma\_t>0$ almost surely for all $t$, then can we
prove
$$P(Y\_1 = c) = 0$$
for all constant $c$?



A non-degenerate martingale
Here's an explicit construction that gives a counterexample. 
For simplicity let $c=0$ (not important).


First, let $\alpha>0$ and consider the probability that a standard Brownian
motion started at 0 hits 0 at some time in the interval
$(\alpha t, \alpha t+t)$.
Then (1) this probability does not depend on $t$ (by Brownian scaling) --
call it $p\_\alpha$; and (2) $p\_\alpha\to 1$ as $\alpha\to 0$
(because with probability 1, the standard Brownian motion hits
0 at some time in the interval $(0,t)$).


Now we'll use this and Borel-Cantelli to show that with positive probability,
we can construct $\sigma$ and a sequence of times $t\_n\uparrow 1$
such that $Y\_{t\_n}=1$ for all $n$.


Let $\alpha\_n$ be some sequence decreasing to 0 quickly enough that
$\sum (1-p\_{\alpha\_n}) <\infty$.


Let $t\_0=0$ and recursively define $t\_1, t\_2, \dots$ as follows.


Given $t\_n<1$ and $Y(t\_n)=0$, let $\sigma\_t=\sqrt{\alpha\_n}$
for $t\in(t\_n, (1+t\_n)/2)$, and let $\sigma\_t=1$
for $t\in((1+t\_n)/2, t\_{n+1})$, where $t\_{n+1}$ is defined by


$t\_{n+1}=\inf \big[ t>(1+t\_n)/2: Y\_t=0 \big]$.


The idea of this definition: given $t\_n<1$,
we divide the remaining time interval $(t\_n, 1)$ into two
halves, and run BM at speed $\alpha\_n$ on the first half
and at speed 1 on the second half, stopping as soon as we hit
0 during the second half.
Since we start at 0,
the probability that we DO hit 0 at some point during the second
half is easily seen to be $p\_{\alpha\_n}$ as defined above.
Hence
$P(t\_{n+1}<1 | t\_n<1)=p\_{\alpha\_n}$.


Now using Borel-Cantelli (and reasoning straightforwardly about independence)
we get that there is positive probability that
$t\_n<1$ for all $n$. In that case also $t\_n\uparrow 1$
(since $1-t\_{n+1}<(1-t\_{n})/2$).
Also $Y\_{t\_n}=0$ for all $n$ by construction.
But the process $Y\_t$ is continuous
(since $\sigma\_t$ is bounded). So then also $Y\_1=0$ as desired.


