If $f:[0;1] \rightarrow \mathbb{R}$ is a continuous function, then we can prove, using Strong law of large numbers on a sequence $X_n$ following the distribution $\mathcal{B}(n,x),x \in [0;1]$, that there exists $Q_n(x)=\sum_{k=0}^nC^k_nf(\dfrac{k}{n})x^k(1-x)^{n-k},x \in \ [0;1],$ such that $$\lim_n \sup_{x \in [0;1]}|Q_n(x)-f(x)|=0.$$ The result remains true if $f$ was continuous on $[\alpha,\beta]$ (considering homeomorphism from $[0;1]$ to $[\alpha,\beta]).$
Looking for a more general case in $\mathbb{R}^d$, if $f$ was continuous on $[0;1]^d,$ then there exists $$Q_n(x_1,...,x_d)=\sum_{0 \leq k_1,...,k_d\leq n}f(\frac{k_1}{n},...,\frac{k_d}{n})\prod_{q=1}^dC_n^{k_q}x_q^{k_q}(1-x_q)^{n-k_q},(x_1...,x_d) \in [0;1]^d,$$ such that $$\lim_n \sup_{x \in [0;1]^d}|Q_n(x)-f(x)|=0.$$ Also, the result remains true on $\prod_{k=1}^d[\alpha_k,\beta_k].$
Do we have the same result on any compact $K$ from $\mathbb{R}^d?$ For a continuous function $f$ on $K$, how to explicit $Q_n$ ?