$\def\d{\mathrm{d}}$Let $f$ be integrable. I want to show there exist two functions $g$ and $h$ that are continuous under a closed interval $[a,b]$ s.t $h\leq f\leq g$ and at the same time$$\int_{a}^{b} (g(x)-h(x)) \,\d x <ε.$$

I know that because $f$ is integrable there exist two steps functions $h\leq f\leq g$ such that$$\int_{a}^{b}g(x) \,\d x - \int_{a}^{b} h(x) \,\d x <ε,$$ but I'm having trouble in the continuity part

My intuition: I'm thinking of "joining" the steps using straight lines in order to have a continuous function. but I have no idea how to formalize it.

Thanks in advance!

  • $\begingroup$ Ah, so you have struggled with the problem. I am happy you posted this question separately. $\endgroup$ – астон вілла олоф мэллбэрг Apr 23 '17 at 8:04
  • $\begingroup$ @Crostul I'll edit the question to show you what I mean $\endgroup$ – user21312 Apr 23 '17 at 8:09
  • $\begingroup$ If $ f $ is not bounded on $(a, b)$ this cant be possible in general $\endgroup$ – ibnAbu Feb 10 '18 at 19:23

I think the picture to have in mind is this:

enter image description here

First, we need to find step functions $f_{\rm red} \leq f$ and $f_{\rm blue} \geq f$ such that $$ \int (f_{\rm blue} - f_{\rm red} )<\frac \epsilon 3.$$ This is possible because $f$ is Riemann integrable.

Next, we want to find "joined-up-step-functions" $f_{\rm green} \leq f_{\rm red}$ and $f_{\rm orange} \geq f_{\rm blue}$ such that $$ \int( f_{\rm red} - f_{\rm green}) < \frac \epsilon 3.$$ $$ \int (f_{\rm orange} - f_{\rm blue}) < \frac \epsilon 3.$$ This is easy to achieve: we just need to make the width of each "triangle" small enough that the areas of the triangles add up to less than $\frac \epsilon 3$. To spell it out: if there are $N$ intervals in the partition, then we should choose the width of each triangle to be less than $2\epsilon /3hN$, where $h$ is the height of the triangle.

  • $\begingroup$ Can this be strengthened so that $g,h$ are polynomials? $\endgroup$ – MathematicsStudent1122 Apr 23 '17 at 9:31
  • 1
    $\begingroup$ If $p$ is a polynomial with $\|f-p\|_\infty<\epsilon/(2(b-a))$, then $p\pm \epsilon/(2(b-a))$ are two polynomials below and above $f$, and the difference between their integrals is $\epsilon$. $\endgroup$ – Fnacool Apr 23 '17 at 11:59
  • $\begingroup$ @Fnacool Yes, that's a really nice argument! :) Although I believe it requires that $f$ is continuous in the first place, so that we can apply the Weierstrass approximation theorem. Nonetheless, we can apply your idea to my $f_{\rm green}$ and $f_{\rm orange}$, using another $\epsilon / 3$ argument - perhaps this is what you had in mind anyway! $\endgroup$ – Kenny Wong Apr 23 '17 at 13:06
  • $\begingroup$ @MathematicsStudent1122 So the answer to your question is YES, as explained by Fnacool. $\endgroup$ – Kenny Wong Apr 23 '17 at 13:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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