I have to prove that for $f \in \mathcal{H}$, \begin{equation} \mathbb{E} \left( \int_0^s f_r \, dW_r \int_0^t f_r \, dW_r \right) = \mathbb{E} \int_0^{s \wedge t} f_r^2 \, dr \end{equation} for any $s,t \in [0, \infty)$, $s \wedge t =\min(s,t)$.
I am not sure how to apply Ito's isometry to a definite integral like this.
I tried to proceed by rewriting $$\int_0^s f_r \, dW_r \int_0^t f_r \, dW_r$ =\int_0^{s \wedge t} 1_{(0, s \wedge t]}f_r \, dW_r \int_0^{s \wedge t} 1_{(0, s \wedge t]}f_r \, dW_r = \left( \int_0^{s \wedge t} 1_{(0, s \wedge t]}f_r \, dW_r \right)^2 $$
and then
\begin{equation} \mathbb{E} \left( \int_0^{s \wedge t} 1_{(0, s \wedge t]}f_r \, dW_r \right)^2 = \int_0^{s \wedge t} \mathbb{E} f_r^2 \, dr \end{equation}
However, I'm not quite sure about this working. Is it correct? If not, where am I going wrong?
Using the equality: $$\int_0^sf_r\,dW_r=\int_0^\infty 1_{]0,s]}(r)f_r\,dW_r,\qquad \text{a.s.} $$ we have \begin{align}\mathsf{E}\Bigl[\int_0^sf_r\,dW_r\int_0^tf_r\,dW_r\Bigr] &=\mathsf{E}\Bigl[\Bigl(\int_0^\infty 1_{]0,s]}(r)f_r\,dW_r\Bigr)\Bigl(\int_0^\infty1_{]0,t]}(r)f_r\,dW_r\Bigr)\Bigr]\\ &=\int_0^\infty1_{]0,s]}(r)f_r1_{]0,t]}(r)f_r\,dr=\int_0^\infty1_{]0,s\wedge t]}(r)f^2_r\,dr\\ &=\int_0^{s\wedge t} f^2_r\,dr. \end{align}