I have met the following proof but it seems too easy to be correct. Anyone to point out the mistakes, if any?
Let $f, f_{n}: [a,b]\to \mathbb{R}$ a continuous functions. If $f_{n}\to f$ in probability, then $E_{p}[f_{n}]\to E_{p}[f]$.
Proof
Let $\epsilon,\delta >0$. Then there exists $n_{0}$ such that for every $n\geq n_{0}$ it is $p(\{ \omega:|f_{n}-f|>\epsilon\})<\delta$. Set $A_{n}=\{ \omega:|f_{n}-f|>\epsilon\}$. Then $|\int (f_{n}-f)dp|\leq \int |f_{n}-f|dp\leq \int |f_{n}-f|X_{A_{n}} + |f_{n}-f|X_{{A_{n}}^{C}}\leq Mp(A_{n})+\epsilon p({A_{n}}^{C})\leq M\delta + \epsilon$
Hence we conclude the convergence.
There are (and must be mistakes), as the statement is wrong.
Let $[a,b] = [0,1]$ with the uniform distribution (hence $\mathbf P$ is Lebesgue measure), define $f := 0$ and $f_n \colon [0,1] \to \mathbf R$ by $$ f_n(x) := \begin{cases} nx & 0 \le x \le \frac 1n\\ 1-n\bigl(x-\tfrac 1n\bigr) & \frac 1n \le x \le \frac 2n \\ 0 & x \ge \frac 2n \end{cases} $$
Then the $f_n$ are continuous, we have $f_n \to f$ pointwise (and hence in probiability), but $$\mathbf E[f_n] = \int_0^1 f_n(x)\, dx = 1 \not\to \int_0^1 f(x)\, dx = 0 = \mathbf E[f] $$ The point where your proof goes wrong is that you assume that you have a common bound for all $\|f_n - f\|_{\sup}$, which can be wrong (if you assume that in addition, your proof [for the then new statement] is fine).