Prove E[Y|X] = f(X) I have a model $Y = f(X) + \epsilon$
where $\epsilon$ is independent of $X$ and $\mathbb{E}[\epsilon]=0, \mathbb{E}\left[\epsilon^2\right]=\sigma^2$.
Show that
$$
f(X)=\mathbb{E}[Y \mid X]
$$
This is my attempt at the proof:
$$
\begin{aligned}
& E[Y \mid X] \\
&= E[f(X)+\varepsilon \mid X] \\
&= E[f(X) \mid X]+E[\varepsilon \mid X] \\
&= E[f(X) \mid X]+E[\varepsilon] \\
&= E[f(X) \mid X]+0 \\
&= E_{Y \mid X}[f(X) \mid X] \quad\quad\quad \quad \quad\ \ (5) \\
&= \sum_{y \mid x} f(X) P(Y|X=y| x) \quad\quad(6) \\
&= f(X) \sum_{y \mid x} P(Y|X=y| x) \quad\quad(7)\\
&= f(X)
\end{aligned}
$$
I am a bit iffy about steps (5) to (6) to (7), please help. Thank you!

$\mathbb{E}[Y|X] = \mathbb{E}[f(X)+\epsilon|X] = \mathbb{E}[f(X)|X]+\mathbb{E}[\epsilon|X] = \mathbb{E}[f(X)|X]+\mathbb{E}[\epsilon] = f(X)+0$
