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Let $g$ be a measurable function on $[0,1]$. Suppose that $g$ is finite almost everywhere and let $\mu$ be the Lebesgue measure. Then for any $\epsilon >0$, there is a polynomial $h$, such that $$ \mu\left(\{x: |g(x)-h(x)|>\epsilon \}\right) < \epsilon.$$

Well, I know that there is a continuous function, say, $f(x)$ such that $|g(x)-f(x)|<\epsilon$ except on a set of measure less than $\epsilon.$ Now, since polynomials are a continuous functions, can I take $h(x)=f(x)$, and thus proving the above? If not, how to I go about proving it?

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No, that wouldn't immediately follow; but, maybe use this – David Mitra Mar 2 '12 at 0:08
@DavidMitra: How do I use it, since the theorem says that for all $x\in [0,1],|g(x)-h(x)|<\epsilon$. – Joe Mar 2 '12 at 0:26
That's fine; it's just more than you need. – David Mitra Mar 2 '12 at 0:27

No, nothing at the outset guarantees that the $f$ you wind up with is a polynomial. The Weierstrass approximation theorem however does.

First choose a continuous function $f$ such that $|g(x)-f(x)|<\epsilon/2$ for all $x\in[0,1]\setminus A$ where the measure of $A$ is less than $\epsilon$. Then use the aforementioned theorem to find a polynomial $p$ such that $|f(x)-p(x)|<\epsilon/2$ for all $x\in[0,1]$. Then $p$ is the desired polynomial.

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So If I get you correctly, I'd have to show that $|g(x)-p(x)|<\epsilon$ and that'll be it? – Joe Mar 2 '12 at 0:59
@Joe Yes. Just use the triangle inequality. – David Mitra Mar 2 '12 at 1:10

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