# A bounded functions between two functions is integrable

Prove that if $\forall \epsilon > 0 \; \exists$ integrable functions $h,g$ at $[a,b]$ s.t $h\leq f \leq g$ and $\int \limits_{a}^{b}(g-f) < \epsilon$ $\implies f$ is integrable.

We've discussed both Riemann and Darboux integrals, and defined integral using Darboux sums. I've tried to use the definition that $f$ is integrable $\iff$ $\forall \epsilon > 0 \; \exists$ step funcions $\phi$ and $\psi$, $\phi \leq f \leq \psi$ s.t $\int \limits_a^b f-\phi < \epsilon$ and $\int \limits_a^b \psi-f < \epsilon$. However I didn't manage to define either $\phi$ or $\psi$ that will do the job. Any ideas?

• There is something wrong in your formulation. $g$ appears too many times. – uniquesolution Apr 15 '17 at 13:13
• @uniquesolution thanks, corrected. – blz Apr 15 '17 at 13:14
• @uniquesolution yes – blz Apr 15 '17 at 13:16
• @uniquesolution you're right, of course. I think they were expecting us to prove it using more basic definitions of integrability, such as Darboux sums, or using step functions. – blz Apr 15 '17 at 13:20

## 1 Answer

Let $P=\{x_0=a,x_1.......x_n=b\}$ a partition of $[a,b]$

Take $\phi(x)=\sum_{k=1}^nl_k\mathbb{X}_{[x_{k-1},x_k]}(x)$ where $l_k=\inf\{f(x):x \in [x_{k-1},x_k]\}$

And $\psi(x)=\sum_{k=1}^nu_k\mathbb{X}_{[x_{k-1},x_k]}(x)$ where $u_k=\sup\{f(x):x \in [x_{k-1},x_k]\}$

These two functions are simple functions..

$\mathbb{X}_A$ is the indicator function of the set $A$

You can continue from here..

• Why, for example, does $int \limits_a^b \psi-f < \epsilon$? – blz Apr 15 '17 at 15:23
• because of the integrability of $f$ ,if yoy assume that $f$ is integrable – Marios Gretsas Apr 15 '17 at 15:24
• the integral of these two functions ,i defined are the lower and upper Darboux sums respectively – Marios Gretsas Apr 15 '17 at 15:25
• The value of each of the indicator functions is the length of the interval – Marios Gretsas Apr 15 '17 at 15:26