Is it true that $f \in L_1([a,b])$ is the uniform limit of polynomials?

Question

Is it true that $f \in L_1([a,b])$ is the uniform limit of polynomials? And why? I know it is the uniform limit on a set take out some finite measurable set but not sure if I can say more. Thanks.

Answer 1

The answer is "sort of"; what you've written (and elaborated on in comments) is not true, but something similar is true. Lusin's theorem tells you that for every $\varepsilon>0$ there exists a compact set $K_{\varepsilon} \subset [a,b]$ such that $m(K_\varepsilon) > b-a-\varepsilon$ and $f$ is continuous on $K_{\varepsilon}$. Weierstrass's theorem gives you a sequence of polynomials which uniformly approximate $f$ on $K_\varepsilon$. You can combine these: take $K_n \subset [a,b]$ with $m(K_n) > b-a-1/n$ and then define $p_n$ so that $\| f - p_n \|_{L^\infty(K_n)} < 1/n$.

The answer is "sort of" instead of "yes" because in general $p_n$ will not converge to $f$ in the sense of $L^{\infty}([a,b])$. To prove this, you need to find an $L^1$ function $f$ such there is no continuous $g$ such that $f=g$ a.e. This is harder than it sounds at first: for example, the indicator function of the rationals is nowhere continuous, but it is a.e. equal to the zero function. As Hans Engler suggested, a function with a jump discontinuity will do the job. Can you prove this fact?

Answer 2

Let us first note that for a continuous function $f : [a,b] \rightarrow \Bbb{R}$, we have

$$ \max_{x \in [a,b]} |f(x)| = \Vert f \Vert_\sup = \Vert f \Vert_\infty = {\rm esssup}_{x \in [a,b]}|f(x)|, $$

as you should verify. The main point here is that $|f|^{-1}((c, \infty))$ is an open set and thus a null-set if and only if it is empty for each $c > 0$.

Now suppose that for $f = \rm{sgn}$, there is a sequence $(f_n)_n$ of continuous functions with $f_n \rightarrow f$ uniformly a.e., i.e. $\Vert f - f_n \Vert_\infty \rightarrow 0$.

This implies in particular that $(f_n)_n$ is Cauchy in $L^\infty$, i.e.

$$ \Vert f_n -f_m \Vert_\sup = \Vert f_n -f_m \Vert_\infty \rightarrow 0 $$

as $n,m \rightarrow \infty$.

By completeness of $C([a,b]), \Vert \cdot \Vert_\sup)$, there is thus some $g \in C([a,b])$ with $\Vert f_n - g \Vert_\sup \rightarrow 0$, i.e. $f_n \rightarrow g$ uniformly. This implies $f = g$ a.e..

But this yields $g \equiv 1$ on $(0,\infty)$ and $g\equiv -1$ on $(-\infty,0)$ (why?), so that $g$ can not be continuous at the origin.