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.
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