Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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
add comment

1 Answer 1

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.