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.
I guess you are reading Measure Theory and Fine Properties of Functions - Evans, and your question is in page25.
Given the "i)" -- "Decomposition of nonnegative measurable functions" in page19 you listed above, it suffices to prove the question. The basic idea of the proof is to use the decomposition to prove the inequality:
$$\int_{*} fd\mu \geq \int^{*} fd\mu$$
but it's a little tricky, here is the proof:
1)$\mu(f=+\infty)>0\ $or$\ \exists n\in N_{+}\ s.t.\ \mu(A_n)=+\infty$
If $\mu(f=+\infty)>0$, any simple function $g$ s.t. $g\geq f, \mu-a.e.$ must have $\mu(g=+\infty) > 0$, and it implies:
$$\int^{*}fd\mu \geq \int_{*}fd\mu \geq (+\infty) * \mu(g=+\infty) = +\infty$$
$$\Rightarrow \int^{*}fd\mu = \int_{*}fd\mu = +\infty$$
therefore f is integrable.
If $\exists n\in N_{+}\ s.t.\ \mu(A_n)=+\infty$, simple function $\frac{1}{n}\chi_{A_n} \leq f$, and therefore:
$$\int^{*}fd\mu \geq \int_{*} fd\mu \geq \int \frac{1}{n}\chi_{A_n} d\mu = \frac{1}{n} * \mu(A_n) = +\infty$$
$$\Rightarrow \int^{*}fd\mu = \int_{*} fd\mu = +\infty$$
which implies $f$ is integrable.
2)$\mu(f=+\infty)=0\ $and$\ \forall n\in N_{+}\ \mu(A_n)<+\infty$
Now we can dismiss the part that $f=+\infty$ and only consider the bounded part of $f$, since the simple funcitons required in the lower and upper integral only require a.e. inequality.
Given the decomposition:
$$f=\sum_{n=1}^{+\infty} \frac{1}{n} \chi_{A_n}$$
we have:
$$\sum_{n=1}^k \frac{1}{n} \chi_{A_n} \leq f \quad \forall k \in N_{+}$$
and $\sum_{n=1}^k \frac{1}{n} \chi_{A_n}$ is simple function by definition, so we have:
$$\begin{align}
\int_{*}fd\mu &\geq \int \sum_{n=1}^k \frac{1}{n} \chi_{A_n} d\mu \\\\
& = \sum_{n=1}^{g(k)} y_n\mu(f^{-1}(B_n)) \quad \forall k \in N_{+}
\end{align}$$
where $B_n = \{x|f(x)=y_n, y_n\in Image(\sum_{n=1}^k \frac{1}{n} \chi_{A_n})\}, g(k)\subseteq N_{+}, g(k)\geq k\ $ ($\{A_n\}$ are not disjoint), and $\mu(B_n)<+\infty$, since $B_n \subseteq \cup_{n=1}^k A_n$ and $\forall n\leq k,\ \mu(A_n)<+\infty$
let $k \rightarrow +\infty$, we have:
$$\int_{*}fd\mu \geq \sum_{0\leq y_n\leq +\infty} y_n\mu(f^{-1}(B_n))$$
where $B_n = \{x|f(x)=y_n, y_n\in Image(\sum_{n=1}^{+\infty} \frac{1}{n} \chi_{A_n})\}$
Since $f < +\infty \ a.e.$ and $f=\sum_{n=1}^{+\infty} \frac{1}{n} \chi_{A_n}$, we have:
$$\forall \epsilon > 0\ \exists N\in N_{+}\ s.t.\ \forall k>N\quad f\leq \epsilon + \sum_{n=1}^k \frac{1}{n}\chi_{A_n}\quad a.e.\ \forall k \in N_{+}$$
therefore:
$$\begin{align} \int^{*}fd\mu &\leq \int \epsilon + \sum_{n=1}^k \frac{1}{n}\chi_{A_n}d\mu \\\\ &= \sum_{n=1}^{g(k)} y_n\mu(f^{-1}(B_n)) + \sum_{n=1}^{g(k)} \epsilon \mu(f^{-1}(B_n)) \quad \forall k \in N_{+} \end{align}$$
let $\epsilon \rightarrow 0^{+}$, we have:
$$\int^{*}fd\mu \leq \sum_{n=1}^{g(k)} y_n\mu(f^{-1}(B_n))\quad \forall k \in N_{+}$$
then let $k\rightarrow +\infty$, we have:
$$\int^{*}fd\mu \leq \sum_{0\leq y_n\leq +\infty} y_n\mu(f^{-1}(B_n))$$
Combine two inequalities above, we have:
$$\int_{*} fd\mu \geq \sum_{0\leq y_n\leq +\infty} y_n\mu(f^{-1}(B_n))\geq \int^{*}fd\mu$$
which implies $f$ is integrable.
I hope you find this proof helpful.
Best Answer
I think you are too focused on trying to use the definition of the integral directly. There are lots of other methods to compute integrals that are much more useful (we just don't use them as the definition itself since it is less clear that they are well-defined).
In particular, any nonnegative measurable function can be integrated by considering it as an increasing limit of simple functions. Think about how you could use that here. A full solution using this idea is hidden below.