Proving Integration by Parts for Brownian Motion and function of bounded variation Looking for someone to help me finish my proof for a version of Integration by Parts. I am trying to prove that if $f$ is a continuous function of Bounded Variation, then $$\int_0^t f(s)dBs = f(t)B_t - \int_0^t B_sdf(s)$$
By definition of the Ito Integral we know that for any partition $\Pi = \{s_0,\ldots s_n\}$ where $x_0=0,x_n=t$ that
$$\lim_{|\Pi|\to0} \sum_{i=1}^n f(s_{i-1}) (B_{s_i}-B_{s_{i-1}})=\int_{0}^tf(s)dBs$$
Where the limit converges in $L^2(P)$. Now Brownian Motion has continuous path almost surely. For any $\omega$ so that $s\to B_s(\omega)$ is continuous, then $B_s(\omega)$ is integrable with respect to $f$ since $f$ is continuous and of bounded variation. Thus $\int_0^t B_s(\omega)df(s)$ exists almost surely. In particular
$$\lim_{|\Pi|\to 0}\sum_{i=1}^n B_{s_i}(\omega)(f(s_i)-f(s_{i-1}))=\int_0^t B_s(\omega)df(s) $$
What I need help is showing that the convergence is also in $L^2(P)$.
If I can show that the convergence is also in $L^2(P)$, then
\begin{align*}\int_0^t f(s)dBs + \int_0^t B_sdf(s) &= \lim_{|\Pi|\to 0}\sum_{i=1}^n f(s_{i-1}) (B_{s_i}-B_{s_{i-1}})+\sum_{i=1}^n B_{s_i}(f(s_i)-f(s_{i-1})) \newline
&= \lim_{|\Pi|\to 0}\sum_{i=1}^n f(s_{i-1}) B_{s_i}-f(s_{i-1})B_{s_{i-1}}+f(s_i)B_{s_i}- f(s_{i-1})B_{s_i}\newline
&= \lim_{|\Pi|\to 0}\sum_{i=1}^n f(s_i) B_{s_i}-f(s_{i-1}) B_{s_{i-1}}\\
&= f(t)B_t - f(0)B_0= f(t)B_t
\end{align*}
where this is the $L^2(P)$ limit.
 A: Consider the a sequence of partitions $\{\Pi_n\}$ on $[0,t]$ where $\Pi_n\subset \Pi_{n+1}$. Here $|\Pi_n| = \max |s_i-s_{i-1}|$ where $\Pi_n = \{s_0, s_1, \ldots s_{m_n}\}$, $0=s_0 < s_1 < \ldots < s_{l_n}=t$ and we assume that $|\Pi_n|\to 0$. Put
$$g_n(\omega)=\sum_{i=1}^{l_n}B_{s_{i}}(\omega)(f(s_i)-f(s_{i-1})) $$
Since Brownian motion is continuous almost surely, then if $t\to B_t(\omega)$ is continuous we see that $g_n(\omega)\to \int_0^t B_s(\omega)df(s)$ since every continuous function is Riemann-Stieltjes integrable with respect to any function of bounded variation.
We now show that $\{g_n\}$ is Cauchy in $L^2(P)$. Assume $n < m$ and without loss of generality, we may assume that for any $s_{i-1},s_i\in \Pi_n$ we have at least one $\tau\in (s_{i-1},s_i) \cap \Pi_m$. We can denote all of the $\tau$ in the following manner $s_{i-1} = \tau_{i,0}< \tau_{i,1} < \ldots < \tau_{i,j_i} = s_{i}$. Then we can write $g_m$ in the following manner
$$g_m(\omega)= \sum_{i=1}^{l_n} \sum_{k=1}^{j_i} B_{\tau_{i,k}}(\omega)(f(\tau_{k,i})-f(\tau_{i,k-1}))$$
We can also rewrite $g_n$ in the following manner
$$g_n(\omega)= \sum_{i=1}^{l_n} \sum_{k=1}^{j_i} B_{s_{i}}(\omega)(f(\tau_{k,i})-f(\tau_{i,k-1})) $$
By our assumption on $\tau$, we know that $i_j \geq 2$ for all $j$ so that $i_j-1\geq 1$ as well. With this in mind, we compute that
$$\begin{align*}\mathbb{E}[(g_n-g_m)^2] &= \mathbb{E}\Big[\Big(\sum_{i=1}^{l_n}\sum_{k=1}^{j_i -1} (B_{s_i}- B_{\tau_{i,k}})(f(\tau_{i,k})-f(\tau_{i,k-1}))\Big)^2\Big]  \\
&= \sum_{r=1}^{l_n} \sum_{u=1}^{l_n} \sum_{w=1}^{j_r-1}\sum^{j_s-1}_{v=1} \mathbb{E}[(B_{s_r}-B_{\tau_{r,w}})(B_{s_u}-B_{\tau_{u,v}})](f(\tau_{r,w})-f(\tau_{r,w-1}))(f(\tau_{u,v})-f(\tau_{u,v-1}))
\end{align*}$$
Now if $u<r$, then we have that $\tau_{u,v}< s_u < \tau_{r,w}<s_r$. Since Brownian Motion has independent increments, then $B_{s_r}-B_{\tau_{r,w}}$ and $B_{s_u}-B_{\tau_{u,v}}$ are independent and both have $0$ mean. Thus $$ \mathbb{E}[(B_{s_r}-B_{\tau_{r,w}})(B_{s_u}-B_{\tau_{u,v}})]=\mathbb{E}[B_{s_r}-B_{\tau_{r,w}}]\mathbb{E}[B_{s_u}-B_{\tau_{u,v}}]=0$$
The same also holds if $r < u$. Hence we can write the previous statement as
$$\begin{align*} \mathbb{E}[(g_n-g_m)^2] &= \sum_{i=1}^{l_n}\sum_{k=1}^{j_i -1} \mathbb{E}[(B_{s_i}- B_{\tau_{i,k}})^2](f(\tau_{i,k})-f(\tau_{i,k-1}))^2\\
&=  \sum_{i=1}^{l_n}\sum_{k=1}^{j_i -1} (s_i-\tau_{i,k})(f(\tau_{i,k})-f(\tau_{i,k-1}))^2\\
&\leq |\Pi_n|\sum_{i=1}^{l_n}\sum_{k=1}^{j_i -1} (f(\tau_{i,k})-f(\tau_{i,k-1}))^2\\
&\leq |\Pi_n|\Big(\sum_{i=1}^{l_n}\sum_{k=1}^{j_i} |f(\tau_{i,k})-f(\tau_{i,k-1})|\Big)^2\\
&\leq |\Pi_n|(V_0^t(f))^2\end{align*}$$
Here $V_0^t(f)$ is the total variation of $f$, since $f$ is assumed to be of bounded variation then $V_0^t(f)<\infty$. Since $|\Pi_n|\to 0$, then we can now observe that $\{g_n\}$ is a Cauchy sequence in $L^2(P)$. Since $L^2(P)$ is a Banach space, then $g_n \to g$ in $L^2(P)$. But we already know that $g_n\to \int_0^t B_sdf(s)$ $a.e$, since we have convergence in both topologies, then the limits must coincide. Hence $g_n \to \int_0^t B_sdf(s)$ in $L^2(P)$.
