Yes. Let $X_t := int^t_0 sigma_s mathrm dW_s$. Due to Theorem V.6 from the book *Stochastic Integration and Differential Equations* (second edition) by P.E. Protter, there is a continuous and adapted process ${tilde X^n_t}_{tin(0;T)}$ such that

$$

tilde X^n_t

=

int^t_0 n cdot big( tilde X^n_s – X_s big) mathrm ds.

$$

Hence we define $tildesigma^n_s := n cdot ( tilde X^n_s – X_s )$, which is adapted and even continuous.

To prove the limit property, we first prove the following:

**Lemma 1.** For all $beta,delta in (0;infty)$, there exists $nu in (0;infty)$ such that

$$

mathbb P bigg( sup_{0 le s le t le min(s+nu,T)} bigg| int^t_s sigma_u mathrm dW_u bigg| le beta bigg)

ge 1 – delta.

$$

**Proof:** We discretize the interval $(0;T)$ and consider events that its increments stay bounded. For all integer $0 le k < N$ and all $alpha in (0;infty)$, we define

$$

A^alpha_{k,N}

:= bigg{ max_{frac{T}{N} k le t le frac{T}{N} (k+1)} bigg| int_{frac{T}{N} k}^t sigma_u mathrm dW_u bigg| ge alpha bigg}.

$$

Due to the Burkholder–Davis–Gundy inequality (since $sigma$ is bounded, say $vertsigmavert le overlinesigma$),

$$

mathbb Ebigg( max_{T/Ncdot k le t le T/Ncdot (k+1)} bigg| int_{T/Ncdot k}^t sigma_u mathrm dW_u bigg|^4 bigg)

le C_4 mathbb Ebigg( bigglangle int_{T/Ncdot k}^cdot sigma_u mathrm dW_u biggrangle_{T/Ncdot (k+1)}^2 bigg)

\=

C_4 mathbb Ebigg( bigg( int_{T/Ncdot k}^{T/Ncdot (k+1)} big(sigma_ubig)^2 mathrm du bigg)^2 bigg)

le C_4 bigg( frac{T} N cdot overlinesigma^2 bigg)^2

= N^{-2} C

$$

and due to the Markov inequality, we obtain

$$

mathbb P big( A^alpha_{k,N} big)

le alpha^{-4} mathbb Ebigg( max_{T/Ncdot k le t le T/Ncdot (k+1)} bigg| int_{T/Ncdot k}^t sigma_u mathrm dW_u bigg|^4 bigg)

le alpha^{-4} N^{-2} C

$$

Now we assume that $omega in Omega backslash bigcup_{k=0}^{N-1} A^alpha_{k,N}$ and assume $s, t in (0;T)$ with $s le t le s + T/N$. Then we can find a $k in {0,ldots,N-1}$ such that $T/Ncdot k le s le T/Ncdot (k+1) le t le T/Ncdot (k+2)$ or $T/Ncdot k le s le t le T/Ncdot (k+1)$. In the first case, we obtain

$$

bigg| bigg(int^t_s sigma_u mathrm dW_ubigg)(omega) bigg|

le bigg| bigg(int_{T/Ncdot (k+1)}^t sigma_u mathrm dW_ubigg)(omega) bigg|

+ bigg| bigg(int_{T/Ncdot k}^{T/Ncdot (k+1)} sigma_u mathrm dW_ubigg)(omega) bigg|

\

quad + bigg| bigg(int_{T/Ncdot k}^s sigma_u mathrm dW_ubigg)(omega) bigg|

le 3 alpha.

$$

In the second case, we get the same result analogously.

Let $omega in Omega backslash bigcup_{k=0}^{N-1} A^alpha_{k,N}$ and $s, t in (0;T)$ with $|s – t| le frac{T}{N}$. Then, $bigg| bigg(int^t_s sigma_u mathrm dW_ubigg)(omega) bigg| leq 3 alpha$ and so

$$

Omega backslash bigcup_{k=0}^{N-1} A^alpha_{k,N}

subseteq bigg{ max_{s,tin (0;T), |s-t| le frac T N} bigg| int^t_s sigma_u mathrm dW_u bigg| le 3 alpha bigg}.

$$

As a result, if $N$ is large enough,

$$

mathbb P bigg( max_{s,tin (0;T), |s-t| le frac T N} bigg| int^t_s tildesigma_u mathrm dW_u bigg| le 3 alpha bigg)

\ge

1 – sum_{k=0}^{N-1} mathbb P big( A^alpha_{k,N} big)

ge 1 – frac{C}{alpha^{4} N^{1}}

ge 1 – delta,

$$

which proves the statement.

Since $tilde X^n$ always moves into the direction of $X$, we also have the following:

**Lemma 2.**

$$

sup_{t in (0;T)} vert tilde X^n_t vert

le

sup_{t in (0;T)} vert X_t vert

$$

Now since the increments of $X$ are bounded on an event of large probability due to Lemma 1, it is also straightforward to prove this:

**Lemma 3.** Let

$$

M^{beta,nu}:=bigg{sup_{0 le s le t le min(s+nu,T)} bigg| int^t_s sigma_u mathrm dW_u bigg| le betabigg}.

$$

Then for all $omegain M$, we have

$$

sup_{tin (0;T)} bigvert tilde X^{beta/nu}_t – X_t bigvert

le 3 beta.

$$

Now we prove the main statement. Let $n:=beta/nu$. Due to the Minkovski inequality,

$$

sqrt{ mathbb Ebigg( int^T_0 big( tilde X^n_s – X_s big)^2 mathrm dt bigg) }

\le

sqrt{ mathbb Ebigg( mathbb 1_{M^{beta,nu}} int^T_0 big( tilde X^n_s – X_s big)^2 mathrm dt bigg) }

+ sqrt{ mathbb Ebigg( mathbb 1_{Omegabackslash M^{beta,nu}} int^T_0 big( tilde X^n_s – X_s big)^2 mathrm dt bigg) }

$$

The first summand can be bound directly by $3 beta sqrt T$ using Lemma 2. The second summand can be bound using Hölder inequality by

$$

mathbb Ebigg( int^T_0 mathbb 1_{Omegabackslash M^{beta,nu}} big( tilde X^n_s – X_s big)^2 mathrm dt bigg)

\le

sqrt{ mathbb Ebigg( int^T_0 mathbb 1_{Omegabackslash M^{beta,nu}} mathrm dt bigg) }

sqrt{ mathbb Ebigg( int^T_0 big( tilde X^n_s – X_s big)^4 mathrm dt bigg) }

=

sqrt T sqrt{1 – mathbb Pbig(M^{beta,nu}big) }

sqrt{ mathbb Ebigg( int^T_0 big( tilde X^n_s – X_s big)^4 mathrm dt bigg) }

$$

The first factor can be made arbitrarily small if choosing $nu$ small enough depending on $beta$ due to Lemma 1, and the second factor is bounded due to the boundedness of $sigma$ and Lemma 2.