Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\frac{h(\xi_{i})}{b-a}$

share|cite|improve this question
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?)

share|cite|improve this answer
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

Your Answer


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.