Let $X$ be an RV, $\mathcal{F}$ a $\sigma$-algebra, $\phi$ the characteristic function of a distribution $\nu$. Assume that for the conditional characteristic function
$$\phi_\mathcal{F}(t) := \mathbb{E}[\exp(itX) \mid \mathcal{F}]$$
holds $$\phi_\mathcal{F}(t) = \phi(t) \quad a.s. \qquad (2)$$
and show that $X$ has distribution $\nu$.
I know that two RVs have the same distribution iff they have the same characteristic function, but I do not see how to use it here. Could you please give me a hint?
Edit: Following Stratis Markou's suggestion I took the conditional expectation of $(2)$:
\begin{align*} \mathbb{E}[\phi_\mathcal{F}(t) \mid \mathcal{F}] &= \mathbb{E}[\phi(t) \mid \mathcal{F}] \\ \Longleftrightarrow \mathbb{E}[\mathbb{E}[\exp(itX) \vert \mathcal{F}] \mid \mathcal{F}] &= \mathbb{E}[\mathbb{E}[\exp(itX)] \mid \mathcal{F}] \\ \end{align*}
In class we have done the tower law/law of total expectation, but when I us this here I only get $\mathbb{E}[\mathbb{E}[\exp(itX) \mid \mathcal{F}]] = \mathbb{E}[\mathbb{E}[\exp(itX)]]$. Am I doing something wrong?
Would it help if you took expectations w.r.t. to the conditioning variable on both sides of the second equation and then applied the law of total expectation?
P.S. I'm a little unsure of what you mean by having a $\sigma$-algebra in the conditioning side in your first equation.