TITLE: Is $x_n(t) = t^n$ for $n\ge 1$ dense in $L^1[0,1]$?
QUESTION [5 upvotes]: My question concerns the following problem

Let $x_n(t) = t^n$ for $n\ge 1$. Is $\mathrm{span}(x_n; n\ge 1)$ dense in $L^1([0,1])$?

By Weierstrass' approximation theorem, it suffices to check whether the constant function $1$ is contained in the $L^1$-closure of the linear span of the $x_n$. 
I think this is not the case, but I haven't found a proof so far.
Some simple observations: I know that if $1$ is contained in the $L^1$-closure, then we can find a sequence of polynomials $p_n\in \mathrm{span}(x_n; n\ge 1)$ such that $p_n\to 1$ in $L^1$, almost everywhere and almost uniformly. Furthermore the sequence $q_n(x) = \int_0^x p_n(t) \, dt$ will converge uniformly to $x$ on $[0,1]$.
I don't see how this helps, though.

REPLY [3 votes]: This is a write up of the solution based on t.b.'s nice idea in the comments:

Claim: Let $x_n(t) = t^n$. Then $\mathrm{span}(x_n; n\ge 1)$ is dense as a subspace of $L^1([0,1])$.

Proof: By monotone convergence, the sequence $x^{1/m}$ converges to $1$ in $L^1([0,1])$. It therefore suffices to check that $x^{1/m} \in \overline{\mathrm{span}(x_n; n\ge 1)}$ for all $m\in \mathbb N$. 
The latter follows from the fact that the sequence of Bernstein polynomials
$$B_k(x) =  \sum_{i=0}^k \binom ki \left(\frac{i}{k}\right)^{1/m}x^i (1-x)^{k-i}$$
converges to $x^{1/m}$ uniformly and is contained in $\mathrm{span}(x_n; n\ge 1)$ (the constant term where $i=0$ vanishes). $\square$