Proving Two Ito Integral Idenities 

*

*Let $f(t)$ be a deterministic function that is of bounded variation on all intervals $[0,s]$ for $s \in \mathbb R$. Also assume that $f$ has compact support in $[0,\infty ]$. Let $B_t$ be the standard Brownian Motion. Show that almost surely $$ \int_0^\infty f(t)\, dB_t=-\int_0^\infty B_t \, df(t) $$


*Now suppose that $f_1$ and $f_2$ are continuous deterministic functions in $t$. Prove that for each $t \ge 0$ almost surely we have that $$ \int_0^tf_2(s)\left[\int_0^sf_1(u) \, dB_u\right] \, ds=\int_0^t\left[\int_s^tf_2(r)\, dr\right]f_1(u)\, dB_u $$

For (1), if $f$ was continuous then I would be able to able to apply Itos formula to the function $g(x,t)=xt$ and then apply this with $g(B_t,f(t))$ but $f$ is not continuous here. However, can I think of the RHS as a Riemann-Stieltjes integral? I feel like my work isn't quite justified. Here is what I did.
First I assumed that the RHS is a Riemann-Stieltjes integral and used the summation by parts formula $$\sum_{k=0}^nf_{k}(g_{k+1}-g_k)=f_{n+1}g_{n+1}-f_0g_0 -\sum_{k=1}^ng_{k+1}(f_{k+1}-f_k)$$
and the compactness of the support of $f$ to get that
$$ \begin{align} \int_0^\infty B_t \, df(t)&= \lim_{n \to \infty}\int_{\frac{1}{n}}^n B_t \, df(t) 
\\
&=\lim_{n \to \infty}\lim_{\Vert \Delta_k \Vert \to 0}\sum_{j=0}^{k-1}B_{t_j}(f(t_{j+1})-f(t_j)) 
\\
&=\lim_{n \to \infty} B_nf(n)-B_{\frac{1}{n}}f(1/n)-\lim_{\Vert \Delta_k \Vert \to 0} \sum_{j=0}^{k-1}f(t_{j+1})(B_{t_{j+1}}-B_{t_j})
\\
& =\lim_{n \to \infty} B_nf(n)-B_{\frac{1}{n}}f(1/n)-\int_{\frac{1}{n}}^nf(t)\, dB_t
\\
&=-\int_0^\infty f(t) \, dB_t
\end{align}$$
Where above $\{\frac{1}{n} = t_0 < t_1 < ... < t_k = n\}$ is a partition of $[\frac{1}{n},n]$ and $\Vert \Delta_k \Vert $ denotes the $\max|t_j-t_{j-1}|$
As for (2) I am unsure on how to proceed. Any hints are appreciated.
 A: Yes, the right-hand side is a Riemann-Stieltjes integral. There is a general result which states that the Riemann-Stieltjes integral $\int_a^b g(t) \, df(t)$ is well-defined if $f$ is of bounded variation and $g$ is continuous. Since Brownian motion has continuous sample paths, this implies that the (Riemann-Stieltjes) integral $\int_a^b B_t \, df(t)$ is well-defined (where $a,b$ are chosen such that the support of $f$ is contained in $(a,b)$).
Regarding your attempt: The idea is correct. You have to be a bit careful about the different modes of convergence. The Itô integral is a priori defined as an $L^2$-limit whereas the Riemann-Stieltjes integral is defined as a pointwise limit. You can wriggle your way out by considering all limits as limits in probability (....since both $L^2$ convergence and pointwise convergence imply convergence in probability).

Note that it follows from the first part of the problem that
$$\int_0^t f(s) \, dB_s = f(t) B_t - \int_0^t B_s \, df(s) \tag{1}$$
for any function $f$ which is of bounded variation. We will use this integration by parts formula several times for the proof of the second part of the problem.

Lemma 1: Let $f,g$ be continuous (deterministic) bounded functions and assume that $g$ is of bounded variation. Then $$\int_0^t f(s) \left( \int_0^s g(r) \, dB_r \right) \, ds = \int_0^t \left( \int_s^t f(r) \, dr \right) g(s) \, dB_s. \tag{2}$$

Proof: Since $g$ is of bounded variation, it follows from $(1)$ that
$$\int_0^s g(r) \, dB_r = g(s) B_s -  \int_0^s B_r \, dg(r).$$
Hence, $$\int_0^t f(s) \left( \int_0^s g(r) \, dB_r \right) \, ds = - \int_0^t f(s) \left( \int_0^s B_r \, dg(r) \right) \, ds + \int_0^t f(s) g(s) B_s \, ds. \tag{3}$$
For the first term on the right-hand side we can apply the classical Fubini theorem to interchange the integrals and this gives
\begin{align*} \int_0^t f(s) \left( \int_0^s B_r \, dg(r) \right) \, ds &=  \int_0^t \left( \int_s^t f(r) \, dr \right) B_r \, dg(r)  \\ &= \int_0^t B_r \, dh(r) \end{align*}
where
$$h(r) := \int_0^r \left( \int_u^t f(s) \, ds \right) \, dg(u).$$
Note that $h$ is a function of bounded variation. Hence, by $(1)$,
$$ \int_0^t f(s) \left( \int_0^s B_r \, dg(r) \right) \, ds =- \int_0^t h(s) \, dB_s + h(t) B_t.$$
Plugging this into $(3)$ we get
$$\int_0^t f(s) \left( \int_0^s g(r) \, dB_r \right) \, ds = \int_0^t h(s) \, dB_s - h(t) B_t + \int_0^t f(s) g(s) B_s \, ds.$$
Applying Fubini's theorem (or the integration by parts formula) it follows that
$$h(s) = \int_0^s g(r) f(r) \, dr + g(s) \int_s^t f(r) \, dr,$$
and so
$$\int_0^t f(s) \left( \int_0^s g(r) \, dB_r \right) \, ds = \int_0^t \left( \int_s^t f(r) \, dr \right) g(s) \, dB_s + X \tag{4} $$
where $$X:= - B_t \int_0^t f(r) g(r) \, dr + \int_0^t f(s) g(s) B_s \, ds + \int_0^t \left( \int_0^s g(r) f(r) \, dr \right) \, dB_s.$$ A final application of $(1)$ yields that $X=0$. Hence, the assertion follows from $(4)$.

Corollary $(2)$ remains valid if $g$ is not necessarily of bounded variation.

For fixed $t>0$ define an approximating sequence
$$g^{(n)}(s) := \sum_{j =0}^{n-1} g(t_j^{(n)}) \cdot 1_{[t_j^{(n)},t_{j+1}^{(n)})}(s), \qquad s \in (0,\infty)$$
where $t_j^{(n)} := t j/n$. Clearly, $g^{(n)}$ has for each fixed $n \in \mathbb{N}$ compact support. Moreover, it follows from the continuity of $g$ that
$$\mathbb{E} \left( \sup_{s \leq 1} \left| \int_0^s g^{(n)}(u) \, dB_u - \int_0^s g(u) \, dB_u \right|^2 \right) \xrightarrow[]{n \to \infty} 0.$$
Consequently, we can find a subsequence of $(g^{(n)})_{n \in \mathbb{N}}$ (which I will denote, in abuse of notation, again by $g^{(n)}$) such that
$$\sup_{s \leq 1} \left| \int_0^s g^{(n)}(u) \, dB_u - \int_0^s g(u) \, dB_u \right| \xrightarrow[]{n \to \infty} 0$$
almost surely. By the dominated convergence theorem this implies,
$$\int_0^t f(s)\left( \int_0^s g(u) \, dB_u \right) \, ds = \lim_{n \to \infty} \int_0^t f(s)\left( \int_0^s g^{(n)}(u) \, dB_u \right) \, ds.$$
Applying the above lemma to $g^{(n)}$ we get
\begin{align*} \int_0^t f(s)\left( \int_0^s g(u) \, dB_u \right) \, ds &= \lim_{n \to \infty} \int_0^t \left( \int_s^t f(r) \, dr \right) g^{(n)}(s) \, dB_s \\ &= \int_0^t \left( \int_s^t f(r) \, dr \right) g(s) \, dB_s. \end{align*}
Remark: The second part of the problem is a simple version of the stochastic Fubini theorem. It tells us that we can interchange the stochastic integral with the Riemann integral (under suitable assumptions on the integrands).
