How to prove a process has a continuous modification? Let $f \in \mathbb{L}^2([0,1])$ and $W(t),~t \in [0,1]$ is standard Brownian motion over $[0,1]$.
Let process $X(t)$ be defined as a Wiener integral of the function $f$ as follows:
$$
X(t)~=~\int_0^t{f(\tau)dW(\tau)},~t \in [0,1]
$$
I was asked a question whether or not it has a continuous modification? I suggest that it has but have no idea how to prove that. I tried to use  Kolmogorov's continuity theorem but failed with that, so I think this case needs special treatment.
UPD. I assume that Wiener integral is introduced as development by continuity of linear mapping $S \rightarrow L^2(\Omega, P)$ $~$($S \subset L^2([0,1])$ is a dense subset of step-functions), defined on $S$ as 
$$
\int fdW(t)~=~\sum_{i\le n}a_i(W(t_i)-W(t_{i-1}))
$$
for
$$
f(t) = \sum_{i\le n}a_i\chi_{[t_{i-1}, t_i]}(t)
$$
 A: Let $f \in L^2([0,1])$. Since the function $$f_n(t) = \begin{cases} n, & f(t)>n, \\ f(t), & f(t) \in [-n,n], \\ -n, & f(t)<-n \end{cases}$$ is deterministic, the stochastic integral
$$X_n(t) := \int_0^t f_n(\tau) \,d W(\tau)$$
is Gaussian. By Itô's formula, the varinace of the Gaussian random variable $X_n(t)-X_n(s)$ equals
$$ \mathbb{E} \left| \int_s^t f_n(\tau) \, dW_{\tau} \right|^2 = \int_s^t f_n(\tau)^2 \, d\tau \leq n^2 |s-t|.$$
As $\mathbb{E}(Y^4) = 3 \sigma^4$ for $Y \sim N(0,\sigma^2)$, we get
$$\mathbb{E}((X_n(s)-X_n(t))^4) \leq 3 n^4 |t-s|^2.$$
Applying Kolmogorov's continuity theorem, we find that $(X_n(t))_{t \geq 0}$ has a modification $(\tilde{X}_n(t))_{t \geq 0}$ with continuous sample paths for each $n \geq 1$. Since, by Doob's maximal inequality,
$$ \mathbb{E} \left( \sup_{t \leq 1} |\tilde{X}_n(t)-X(t)|^2 \right) = \mathbb{E} \left( \sup_{t \leq 1} |X_n(t)-X(t)|^2 \right) \leq 4 \int_0^1 (f(\tau)-f_n(\tau))^2 \, d\tau$$
it follows from the dominated convergence theorem that the right-hand side converges to $0$ as $n \to \infty$. Choosing an almost surely convergent subsequence, we find
$$\sup_{t \leq 1} |X_{n_k}(t)-X(t)|^2\xrightarrow[]{k \to \infty} 0$$
and therefore $(X(t))_{t \in [0,1]}$ has continuous sample paths with probability $1$ (as a uniform limit of processes with a.s. continuous sample paths).
Remark: Typically the continuity is shown when constructing Itô's integral; the idea is to take a sequence of simple processes $(f_n)_n$ approximating $f$ and then to use that
$$\int_0^t f(\tau) \, dW(\tau) = L^2-\lim_{n \to \infty} \int_0^t f_n(\tau) \, dW(\tau)$$
converges uniformly in $t \in [0,T]$. Since $\int_0^{\bullet} f_n(\tau) \, dW(\tau)$ is continuous by the very definition of the Itô integral, this implies that $\int_0^{\bullet} f(\tau) \, dW(\tau)$ has almost surely continuous sample paths. From my point of view, this is much more straight forward than the above reasoning. 
A: We need to recall the definition of stochastic integral.
If your stochastic integral is defined in Kunita-Watanabe sense, it has continuous sample paths by definition.
