Suppose $X\_m\sim p\_m(x)$ is a discrete distribution on $[0,1]$ where the value takes multipliers of $\frac{1}{m}$ (e.g., $p\_m(x=\frac{k}{m})=\frac{1}{m+1})$. Suppose $p(x)=\lim\limits\_{m\rightarrow\infty}p\_m(x)$, that is, $p\_m$ converge weakly to $p$ on $[0,1]$. (e.g., $p(x)$ is the uniform distrubtion on $[0,1]$).


I wonder the following generalization on bernstein polynomials is true. $\lim\limits\_{n\rightarrow\infty}\sum\limits\_{l}p\_n{(x=\frac{l}{n})}\cdot\left(\sum\limits\_{k=0}^n\binom{n}{k}x^k(1-x)^{n-k}u(\frac{k}{n})\right)=u\left(\int\limits\_{0}^1 xp(x)dx\right)$ for any continuous $u$.



On probabilistic extension for Bernstein polynomials
The limit is not $u(\int\_0^1 x p(x)\,dx)$ but rather $\int\_0^1 u(x) p(x)\,dx$.


I prefer probabilistic notation, so let $X\_n \sim p\_n$ and $X \sim p$. We are then supposing that $X\_n \Rightarrow X$ in distribution. Let $Y\_n$ have conditional distribution $\mathrm{Bin}(n, X\_n)$ given $X\_n$. The question is then about $\lim\_{n \to \infty} E[u(Y\_n / n)]$. Your conjecture is that it equals $u(E[X])$ but it actually equals $E[u(X)]$.


Fix $\epsilon > 0$. By uniform continuity there exists $\delta > 0$ such that if $|s-t| < \delta$ then $|u(s)-u(t) < \epsilon|$. Now by Chebyshev's inequality, we have
$$\begin{align\*} P\left(\left|\frac{Y\_n}{n} - X\_n\right| \ge \delta\right) &\le \delta^{-2} \operatorname{Var}\left( \frac{Y\_n}{n} - X\_n\right) \\
&= \delta^{-2} E\left[\operatorname{Var}\left(\frac{Y\_n}{n} \mid X\_n \right)\right] \\
&= \delta^{-2} n^{-1} E[X\_n (1-X\_n)] \\
&\le \delta^{-2} n^{-1}.\end{align\*}$$ 
Hence
$P\left(|u(\frac{Y\_n}{n}) - u(X\_n)| \ge \epsilon\right) \le \delta^{-2} n^{-1}$, so $|u(\frac{Y\_n}{n}) - u(X\_n)| \to 0$ in probability. By dominated convergence (since $u$ is bounded), $E[u(\frac{Y\_n}{n})] - E[u(X\_n)] \to 0$ also. But since $X\_n \Rightarrow X$, we have $E[u(X\_n)] \to E[u(X)]$.


