Suppose $x_0$ is the only discontinuous point of $f$ in $[0,1]$ , $f(x_0-)$ and $f(x_0+)$ exist. $$B_n(f)(x):=\sum_{k=0}^{n} f(\frac{k}{n}) \binom{n}{k} x^k (1-x)^{n-k}=\sum_{k=0}^n f(\frac{k}{n})p_{n,k}(x)$$ I want to prove that $\lim_{n \rightarrow \infty} B_n(f)(x_0)=\frac{f(x_0+)+f(x_0-)}{2}$
Here is what I have done.(It comes from one method of proving the Weierstrass approximating theorem)
$X_n:=\{k \in \{0,1\cdots n\}|x_0\le \frac{k}{n}\le x_0+ \frac{1}{n^{\beta}}\} , $ $Y_n:=\{k \in \{0,1\cdots n\}|x_0-\frac{1}{n^{\beta}}\le \frac{k}{n} < x_0\} , $ $Z_n:=\{k \in \{0,1\cdots n\}||\frac{k}{n}-x_0| > \frac{1}{n^{\beta}}\}$ where $\beta \in (0,1)$ is not decided yet.
$$B_n(f)(x_0)-\frac{f(x_0+)+f(x_0-)}{2} = \sum_{k=0}^n (f(\frac{k}{n})-\frac{f(x_0+)+f(x_0-)}{2})p_{n,k}(x_0)= \sum_{k \in X_n} (f(\frac{k}{n})-f(x_0+))p_{n,k}(x_0)+\sum_{k \in Y_n} (f(\frac{k}{n})-f(x_0-))p_{n,k}(x_0)+\frac{f(x_0+)-f(x_0-)}{2}(\sum_{k \in X_n}p_{n,k}(x_0)-\sum_{k \in Y_n}p_{n,k}(x_0))+\sum_{k\in Z_n} (f(\frac{k}{n})-\frac{f(x_0+)+f(x_0-)}{2})p_{n,k}(x_0) $$
For the first two terms, we have$|f(\frac{k}{n})-f(x_0+)|\le\omega(\frac{1}{n^{\beta}})$ when $k\in X_n$ , $|f(\frac{k}{n})-f(x_0-)|\le \omega(\frac{1}{n^{\beta}})$ when $k \in Y_n$,where $\omega(0+)=0$
For the last term, when $k \in Z_n$,we have $$1\le n^{2\beta}(\frac{k}{n}-x_0)^2=n^{2\beta-2}(k-nx_0)^2 , $$so it is bounded by $$2Mn^{2\beta-2}\sum_{k \in Z_n}(k-nx_0)^2p_{n,k} \le 2Mn^{2\beta-1}x_0(1-x_0) \le M n^{2\beta-1}$$
$M=sup_{x \in [0,1]} |f(x)| . $The equality $\sum_{k=0}^n (k-nx)^2p_{n,k}(x)=nx(1-x)$ is used.
So we take $\beta \in (0,\frac{1}{2})$ .
As for the second term, it goes to $0$ intuitively, but I failed to give a estimation . Could someone help me ?