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Here is my homework. $f\in L([a,b]),\epsilon>0$.Prove that exist a step function $S(x)$ such that $$\int_{a}^{b}|f(x)-S(x)|dx<\epsilon$$ My method:

Assume that $f$ is non-negative, or we will discuss $f_{+}$ and $f_{-}$ of $f$.

Since we can find a continuous function $h(x)$ s.t. $$\int_{a}^{b}|f(x)-h(x)|dx<\epsilon$$ So,we just need to find $S(x)$,s.t. $$\int_{a}^{b}|S(x)-h(x)|dx<\epsilon$$ Obviously,$h(x)$ is Riemann integral,for the $\epsilon$ above, there exist a partition of $[a,b]$$\, a = x_0 < x_1 < x_2 < \cdots < x_n = b\,$,$\exists\xi_{i}\in(x_{i-1},x_i)$,s.t. $$|\int_{a}^{b}h(x)dx-\sum_{i=1}^{n}h(\xi_{i})(x_{i}-x_{i-1})|<\epsilon$$ That is $$\int_{a}^{b}|h(x)-\frac{1}{b-a}\sum_{i=1}^{n}h(\xi_{i})(x_{i}-x_{i-1})|dx<\epsilon$$ Then we find the step function $S(x)$ $$S(x)=\frac{1}{b-a}\sum_{i=1}^{n}h(\xi_{i})(x_{i}-x_{i-1})$$ At the interval $(x_{i-1},x_i)$,we have find a constant $\frac{h(\xi_{i})}{b-a}$

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up vote 1 down vote accepted

Let's expand a bit what you've done. We have that

$|\int_{a}^b h(x) dx = \displaystyle \sum_{i=1}^{n} h(\xi_i)(x_i - x_{i-1})| < \epsilon$.

By noting that $\frac{1}{b-a}\int_{a}^b dx = 1$, we can move the sum underneath the integral and see that

$|\int_{a}^b h(x) - \sum_{i=1}^{n} h(\xi_i)(x_i - x_{i-1})dx| < \epsilon$.

This isn't quite the same as what you have. You can't move the absolute value underneath the integral. However, the following two devices will allow for this:

1) We may assume h is nonnegative (why?)

2) We may approximate h from below (why?)

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Yeah,I should point out that $h(x)$ should be nonnegative,but why do we need to approximate h from below ? –  user39843 Nov 22 '12 at 3:30
    
Sorry, in this case it's above. What you're trying to show is that there's a step function $S(x)$ which satisfies $\int_{a}^b |S(x) - h(x)| dx \le \epsilon$. What you have so far is that $|\int_a^b S(x) - h(x) dx| \le \epsilon$. However, these are not the same thing. If additionally $h(x) \le S(x)$, then $\int_a^b S(x) - h(x) dx = \int_a^b |S(x) - h(x)| dx$, because $S - h \ge 0$. You should also be careful about your usage of two different integrals here. You're using a Riemann integral to approximate h and a Lebesgue integral (I assume) for f. Do you know that these are the same? –  anonymous Nov 22 '12 at 21:28
    
You're right. I use the inf finally. –  user39843 Dec 9 '12 at 13:35
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