I am having trouble understanding one of the steps in the proof of the following lemma.
Let $X$ be a random $d$-vector and $\mathcal{A}$ a sub-$\sigma$-algebra on the probability space $(\Omega,\mathcal{F},P)$. Let $$\phi(\theta) = \int_{\mathbb{R}^d}e^{i\theta^\top x }\mu(dx)$$ be the characteristic function of a probability distribution $\mu$. If $$E[e^{i\theta^\top X}\mathbb{1}_A] = \phi(\theta)P(A)$$ for all $\theta \in \mathbb{R}^d$ and $A \in \mathcal{A}$, then $X$ has distribution $\mu$ and is independent of $\mathcal{A}$.
The step in the proof that I don't understand is the following. We fix $A \in \mathcal{A}$ and define the probability measure $\nu_A$ by $$\nu_A(B) = \frac{1}{P(A)}E[1_B(X)1_A] \quad B\in\mathcal{B}_{\mathbb{R}^d}$$ Then the text reads: "By hypothesis, the characteristic function of $\nu_A$ is $\phi$". I don't see why that is true. How does one compute the characteristic function of $\nu_A$ based on the way it is defined?
Claim: for any measurable function $f : \mathbb{R}^d \to \mathbb{R}$, we have $\int f\,d\nu_A = \frac{1}{P(A)} E[f(X) 1_A].$ You have already proved this for $f$ of the form $f = 1_B$. Then prove it for simple functions $f$ using linearity, for nonnegative $f$ using monotone convergence, and then for general $f$. Finally apply it with $f(x) = e^{i \theta^T x}$.