Lebesgue integrable implies Riemann integrable? I'm self studying measure theory by Bartle's book and there he defined integrability for non-negative functions as follow

Definition: Let $f$ a non-negative measure function, then the integral of $f$ is
$$\int f = \sup \left\{ \int \phi : \ \phi \leq f  \right\},$$
where $\phi$ is a simple function.

It's clear that this definition is the same of lower integral for Riemann integral. I think a definition equivalent to upper integral to Riemann integral is
$$\int f = \inf \left\{ \int \phi : \ \phi \geq f  \right\},$$
where $\phi$ is a simple function.
My doubt is if the definition of integral by Bartle's book implies the definition of Riemann integrable, i. e., is the following true?$$\sup \left\{ \int \phi : \ \phi \leq f  \right\} = \inf \left\{ \int \phi : \ \phi \geq f  \right\}$$
Thanks in advance!
 A: 
It's clear that this definition is the same of lower integral for Riemann integral (...)

That is false. As a counter-example, the function $\mathbf{1}_{\mathbb{[0,1] \backslash Q}}: [0,1] \to \mathbb{R}$ has lower integral for Riemann integral equal to $0$, and "lower" integral according to the Lebesgue definition equal to $1$. The point, as mentioned by Ian at the comments, is that not every simple function is a step function: the function above  being an example.
The function above is also an example of one which is Lebesgue integrable but not Riemann integrable, so the statement as it is in the title is not true.
However, the statement
$$\sup \left\{ \int \phi : \ \phi \leq f  \right\} = \inf \left\{ \int \phi : \ \phi \geq f  \right\}$$
is true, if $f$ is a non-negative bounded function which is not zero only on a finite measure set.* To see this, it suffices to show a sequence of simple functions $\phi_n \geq f$ such that $\lim \int \phi_n =\int f$.
Pick $M \mathbf{1}_E \geq f$. Now, we then have $M\mathbf{1}_E - f \geq 0.$ It follows that there is an increasing sequence $s_n$ of simple functions such that $s_n \to M\mathbf{1}_E-f$ and $s_n \leq M\mathbf{1}_E-f.$ By the monotone convergence theorem, $\int s_n \to \int M\mathbf{1}_E -\int f$. 
We have that $f \leq M\mathbf{1}_E-s_n$, so that $\phi_n:=M\mathbf{1}_E-s_n$ is a sequence of simple functions satisfying what we want, since
$$\int \phi_n=\int M\mathbf{1}_E-\int s_n \to\int f.$$
*If $f$ doesn't satisfy those hypotheses (i.e., bounded and not zero only on a finite measure set), the right side is always infinity so the question is a little irrelevant.
