A non-degenerate martingale 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: 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.
A: I have not thought about such questions for a while, but I do not see an immediate mistake in the following reasoning:
Let $Y_t=E[\mathrm{sign}(W_1)| \mathcal{F}_t]$. Then it is a bounded martingale, and it also has a continuous modification since $Y_t=E[\mathrm{sign}(W_1) | W_t]$ and $E[\mathrm{sign}(W_1) | W_t=x]$ depends on $t$ and $x$ continuously. Therefore, it admits a representation via stochastic integral, but $P(Y_t=\pm1 )=1/2$.
This does not contradict the stochastic representation since $\sigma_t$ in it becomes increasingly large if $t$ is close to 1, but the process is far from $\pm1$. This diffusion pushes the process closer to the boundaries of $[-1,1]$.
Is this true, or am I missing something?
A: Another counter-example: Let $f:R\to R$ be a non-constant $C^2$ funtion  with $f, f', f''$ all bounded, and such that (i) $f$ vanishes in a non-empty open interval $I$, (ii) $f'>0$ outside the closure of $I$,  and (iii) $E[f(W_1)]=0$. Consider the random variable $Y_1:= f(W_1)$. Clearly $P[Y_1=0]= P[W_1\in I]>0$. The process 
$$
\sigma_t(\omega):=h(t,W_t(\omega)),
$$
where 
$$
h(t,x) :=\cases{\int_R {1\over\sqrt{2\pi(1-t)}} \exp[-(y-x)^2/2(1-t)] f'(y) dy,& 0\le t<1,\cr
0,&t\ge 1,\cr}
$$
is predictable and bounded, and $\sigma_t(\omega)>0$ fo all $t\in[0,1)$ and all $\omega\in \Omega$. Moreover, by Ito's formula,
$$
Y_1=\int_0^1 \sigma_t dW_t,
$$
almost surely. In fact,
$$
E[f(W_1)|{\mathcal F}_s]=\int_0^s \sigma_t\,dW_t,\qquad\forall s>0,
$$
almost surely.
