Ito isometry under characteristic function

Question

In general for a stochastic integral of the form $\int_0^1 f(s) dB_s$ (when it makes sense) satisfies the Ito isometry: $E[(\int_0^1 f(s)dB_s )^2] = E[ \int_0^1 f^2(s)ds ].$ If $\Omega$ is our probability space, for a measurable set subset $A \subset \Omega,$ does it hold that $E[ ( \int_0^1 f(s)dB_s )^2 1_A ] = E[ \int_0^1 f^2(s)ds 1_A ]$?. I've checked several books but could not find a reference about this result, any help is appreciated.

Answer 1

Take $f(s) = 1$ for all $s\in[0,1]$ and $A = \{|W_1| < \frac12\}$, then $$ \Bbb E\left[\left(\int_0^1 f^2(s)\mathrm ds\right) 1_A\right] = \Bbb E[1_A] = \Bbb P(A) $$ but $$ \Bbb E\left[\left(\int_0^1 f(s)\mathrm dW_s\right)^2 1_A\right] = \Bbb E\left[W_1^2 \cdot 1_A\right] \leq \frac14\Bbb P(A). $$