Proof of integrability of function through step functions

Question

Prove that function $f(x)$ is integrable if and only if for all $ε > 0$, there exist step functions $s, t$ (defined on the same bounds as $f$) such that $s ≤ f ≤ t$ and $$∫(t-s)(x)dx < ε. \quad \square$$

Answer 1

Let $[a,b]$ be the bounded intervall on which $f$ is defined and bounded and $\mathcal E$ the set of step functions on $[a,b]$.

$\Longrightarrow$ : Assume $f$ is Riemann integrable, we have $$ I(f)_-:= \sup_{\phi \in \mathcal E, \phi \leq f} I(\phi) = \inf_{\psi \in \mathcal E,\psi \geq f} I(\psi) =: I(f)_+ $$ the common value being a real number. Then by definition of the sup/inf, forall $\epsilon>0$, there exists $\phi,\psi \in \mathcal E$ such that $\phi \leq f \leq \psi$ and $$I(\psi) - \epsilon < I(f)_+ = I(f)_-< I(\phi) + \epsilon.$$ Then by linearity you get $$ I(\psi-\phi) = I(\psi)-I(\phi) < 2 \epsilon. $$ $\Longleftarrow$ : Assume that forall $\epsilon > 0$ there exists $\phi,\psi \in \mathcal E$ such that $\phi \leq f \leq \psi$ and $I(\psi - \phi) < \epsilon$. We show that $I(f)_+ = I(f)_-$. We have one inequality for free thanks to the monotony of the Riemann integral over simple functions : forall $\phi,\psi \in \mathcal E$ such that $\phi \leq f \leq \psi$ we have $$ I(\phi) \leq I(\psi) $$ so $$ I(f)_ -\leq I(f)_+. $$ For the other inequality, notice that by assumption you have that forall $\epsilon > 0$, there exists $\phi,\psi \in \mathcal E$ such that $\phi \leq f \leq \psi$ and $I(\psi - \phi) < \epsilon$, so $$ I(f)_+ \leq I(\psi) < I(\phi) + \epsilon \leq I(f)_- + \epsilon $$ and taking $\epsilon$ to 0 you are done.