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Assume $f$ is integrable over $[a,b]$ and $\epsilon > 0$. Show that there is a continuous function $h$ over $[a,b]$ for which $h(x) \leq f(x)$ for all $x \in [a,b]$ and $\displaystyle \int_{a}^b (f(x)-h(x))dx < \epsilon$.

This is a follow-up question to this. We can't use the definition of Remannian integrability as in that question since we need $h(x)$ to be continuous. How should we use the fact that $f$ is integral to find a continuous function $h(x)$?

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Puzzled417
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1 Answers1

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Riemann integrability directly gives you a step function $s \leq f$ with $\int_a^b (f(x)-s(x)) dx < \epsilon/2$. You should work out why this is. (Hint: $\int_a^b s(x) dx$ is a Riemann sum for $\int_a^b f(x) dx$.)

You now wish to argue that, given a step function $s$, there is a continuous function $h \leq s$ with $\int_a^b (s(x)-h(x)) dx<\epsilon/2$. You can achieve this by making $h$ piecewise linear. Before working on the general case, try it in the particular case where $a=-1,b=1$ and $s$ is the unit step function at zero.

Once you have this, $f(x)-h(x)=f(x)-s(x)+s(x)-h(x)$ and so the result follows by linearity of Riemann integration.

Ian
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  • If I let $g(x) = \inf{f(x) | x \in [x_{i-1},x+i]}$ how do I make this continuous? – Puzzled417 Apr 25 '16 at 03:04
  • @Puzzled417 I hinted at how to do it: you want to find a piecewise linear function which is less than what you called $g$ (I called it $s$) but close to it. Here's a further hint: in the special case I pointed out, you can use $h(x)=\begin{cases} 0 & x<0 \ x/(\epsilon/2) & 0 \leq x < \epsilon/2 \ 1 & x \geq \epsilon/2 \end{cases}.$ This is exactly equal to $s$ except on $(0,\epsilon/2)$ and deviates from $s$ by at most $1$ on that interval, so the integrals differ by less than $\epsilon/2$. And it is not hard to see that $h$ is continuous. – Ian Apr 25 '16 at 11:46
  • @Puzzled417 It is straightforward but a bit tedious to adapt this to a general step function (you have to be a bit careful about how many jumps you have, how big the jumps are, and how close together they are). – Ian Apr 25 '16 at 11:48
  • I am not sure how to pick $s(x)$ then since you want the lower Riemann sum. I said $s(x) = \text{inf}{f(x)|x \in [x_{i-1},x_i]}$. – Puzzled417 Apr 26 '16 at 00:01
  • @Puzzled417 Yes, that is the correct $s(x)$ (except for the fact that you shouldn't re-use the letter $x$ like that). You then approximate that by a continuous function $h(x)$. – Ian Apr 26 '16 at 00:10
  • I was thinking of just connecting the endpoints of each $f(x_i)$ to $f(x_{i+1})$ and then making it $f(x_0)$ for $x < x_0$ and $f(x_n)$ for $x > x_n$. That doesn't change the area given by the step function integral. – Puzzled417 Apr 26 '16 at 00:30
  • @Puzzled417 What do you mean "connect"? Do you mean that the result is piecewise linear? If so, that does indeed change the integral, but you can make it only a very slight effect. – Ian Apr 26 '16 at 00:46
  • The line segments are vertical, so I don't see how it changes the integral. – user19405892 Apr 26 '16 at 02:20
  • @user19405892 What? The graph of a function can't contain any vertical line segments. – Ian Apr 26 '16 at 02:22
  • We can make it limit a line segment. – user19405892 Apr 26 '16 at 02:44
  • @user19405892 Yes but proving that rigorously is exactly what I said. – Ian Apr 26 '16 at 03:01