Can we prove Weierstrass Approximation using Strong Law of Large Numbers? I am looking at Question 17 of the Exercises in these notes (pp. 315), which is looking for a proof of the Weierstrass Approximation Theorem using probabilistic methods.  I have only been able to the prove point wise convergence until now.  I am not sure if the answer to the question is yes or no.  If Yes (can somebody prove it or give a slight hint).

Weierstrass Approximation Theorem. Suppose  $f$  is a continuous real-valued function defined on the real interval $[a, b]$. For every $\varepsilon > 0$, there exists a polynomial $p(x)$ such that for all $x$ in $[a, b]$, we have $| f (x) − p(x)| < \varepsilon$, or equivalently, the supremum norm $|| f  − p|| < \varepsilon$.
 A: See also this question; the proof is sketched in the related comments by @cardinal.
Without loss of generality we can assume that the interval is $[0,
\,1]$. Consider the following Bernstein's polynomial
$$ B_n(x) :=
\sum_{k= 0}^n f(k/n) { n \choose k} x^k (1 - x)^{n-k}
$$
which will provide an approximation of $f(x)$: we can prove that
$B_n(x)$ tends to $f(x)$ uniformly in $x$ for large $n$. Although this is
standard mathematical analysis, we can use the formalism of
probability theory, writing some sums as an expectation or a variance
which makes things simpler. We can indeed write $B_n(x) =
\mathbb{E}[f(S_n/n)]$ where $S_n$ follows the binomial distribution
with size $n$ and probability $x$. We know by the law of large numbers
that for $n$ large $S_n/n$ is close to $x$, hence that
$\mathbb{E}[f(S_n /n)]$ is close to $\mathbb{E}[f(x)]= f(x)$. A little
bit of extra work is needed to prove the wanted uniform convergence.
Define the r.vs $D_n := | f(x) - f(S_n/n)|$ and $E_n := |x -S_n/n |$
the distribution of which depends on $n$ and $x$. For any $x$ in the
interval $[0,\,1]$
$$
    \left| f(x) - B_n(x) \right| =
    \left| \, \mathbb{E}[f(x)] - \mathbb{E}[f(S_n/n)] \, \right|
    \leqslant \mathbb{E}[D_n].   
$$
Now for $\delta >0$ we have
$$
   \mathbb{E}[D_n] =
   \mathbb{E}[D_n \vert\, E_n \leqslant \delta]
   \, \text{Pr}\{ E_n \leqslant \delta \}
   +
    \mathbb{E}\left[D_n \vert\, E_n > \delta \right]
   \, \text{Pr}\{E_n > \delta \} =: a + b,
$$
and we consider the two terms $a$ and $b$ separately. For the first term
$$
a \leqslant \mathbb{E}[D_n \vert\, E_n \leqslant \delta]
  \leqslant \omega_f(\delta)
$$
where $\omega_f(\delta)$ is the modulus of continuity of $f$ on
$[0,\,1]$ defined as the supremum of $|f(u) - f(v)|$ for all $u$, $v$ with $|u - v| \leqslant \delta$.  In the second term the expectation is $\leqslant
2B$ where $B:=\sup_x |f(x)|= \|f\|_{\infty}$ so
$$ b \leqslant 2B \, \text{Pr}\{E_n > \delta \} \leqslant 2B \,
\frac{x(1-x)}{n \delta^2} \leqslant \frac{2B}{n \delta^2} $$
where Chebyshev's inequality was used with $\text{Var}[S_n /n] = x(1-x) /n$.
If $\epsilon >0$ the inequality $| f(x) - B_n(x) | < \epsilon$
can be seen to hold for all $x$ when $n$ is large enough.
Indeed we can choose $\delta > 0$ such that
$\omega_f(\delta) < \epsilon /2$ due to the uniform continuity of $f(x)$
on the compact interval $[0,\,1]$. Then for $n$ large enough the
second term is $b < \epsilon /2$  hence
$| f(x) - B_n(x) | < \epsilon$ as claimed.
As a final remark, using Bernstein's polynomials is in practice a
very inefficient method to approximate a given function.
A: Weierstrass Approximation Theorem: Suppose $f$ is a continuous real-valued function defined on the real interval $[a,b]$.  For every $\varepsilon > 0$ there exists a polynomial $p$ such that for all $x \in [a, b]$ we have $|f(x)−p(x)| < \varepsilon$ (or equivalently, the supremum norm $||f−p|| < \varepsilon$).

The notes ask you to prove the result on the unit interval, so let's take $a=0$ and $b=1$ in the Weierstrass approximation theorem.  (The proof can easily be generalised to the broader case.)  The polynomial $p$ in the theorem is constructed by taking coin tosses $X_1,X_2,X_3,... \sim \text{IID Bern}(x)$ leading to $S_n \sim \text{Bin}(n,x)$.  We use this statistic to construct the polynomial:
$$\begin{align}
p(x) 
&\equiv \mathbb{E}(f(S_n/n)) \\[6pt]
&= \sum_{s=0}^n f(s/n) \text{Bin}(s|n,x) \\[6pt]
&= \sum_{s=0}^n f(s/n) {n \choose s} x^{s} (1-x)^{n-x}. \\[6pt]
\end{align}$$
The results in the linked question show that $p(x) = \mathbb{E}(f(S_n/n)) \rightarrow f(x)$ for all $0 \leqslant x \leqslant 1$, which is then used to prove the approximation theorem.  In order to establish the approximation theorem you need to show uniform convergence, which is possible to prove as a result of the strong law of large numbers.
A: Let Sn satisfies strong law, then it satisfies weak law and then proceed with the standard weak law proof you mentioned above
