In the textbook the authors define the integral via cauchy sequences of simple functions: $$S_n\to F:\quad\int F\mathrm{d}\mu:=\lim_n\int S_n\mathrm{d}\mu\quad\left(\int\|S_m-S_n\|\mathrm{d}\mu\to0\right)$$ Now, how come that this is really the usual Lebesgue integral for positive functions: $$f\geq0:\quad\int f\mathrm{d}\mu:=\lim_n\int s_n\mathrm{d}\mu=\sup_{s\leq f}\int s\mathrm{d}\mu$$
(I tried wiggling around with monotone and dominated convergence but couldn't get ahead.)
On
Ok, let us establish both directions. Let us denote the Amann-Escher integral by $\int_A$ and the usual integral simply by $\int$. As noted in the comments, $\int_A = \int$ on simple functions.
"$\Rightarrow$": Let $f : X \to [0,\infty)$ be Lebesgue integrable with the usual definition. Then there is an increasing sequence $f_n$ of simple functions with $0 \leq f_n \to f$. We then have $\int_A f_n \, d\mu \leq \int_A f_{n+1} \, d\mu = \int f_{n+1} \, d\mu \leq \int f \, d\mu =: C$, so that the sequence $(\int_A f_n \, d\mu)_n$ is increasing and bounded, hence convergent to some limit $$L = \lim_n \int_A f_n \,d\mu = \lim_n \int f_n \, d\mu = \int f \, d\mu.$$
where the last equality is due to monotone convergence for the usual integral.
Then note (for $n \leq m$):
$$ \int_A |f_n - f_m| \, d\mu = \int_A f_m - f_n \, d\mu \to L - L = 0\text{ as } n,m\to \infty. $$
Hence, $(f_n)_n$ is $L^1$-Cauchy with $f_n \to f$ pointwise. In particular, $f$ is Amann&Escher-integrable :). Hence (by definition)
$$ \int_A f \,d\mu = \lim_n \int_A f_n \, d\mu = \lim_n \int f_n \, d\mu = \int f \, d\mu, $$ where the last step is due to monotone convergence for the usual integral again.
"$\Leftarrow$": Let $f : X \to [0,\infty)$ be Lebesgue integrable w.r.t. the Amann-Escher definition. There is a sequence $(f_n)_n$ as above.
Again, we have $\int f_n \,d \mu = \int_A f_n \, d\mu \leq \int_A f \, d\mu$ for all $n$ (that $f \leq g $ implies $\int_A f \, d\mu \leq \int_A g \, d\mu$ for integrable $f,g$ is easy to verify, simply note that $f \geq 0$ implies $\int_A f \,d\mu \geq 0$ and that the integral is linear).
You can now repeat essentially the same steps as above to see that $(f_n)_n$ are L^1-Cauchy with $f_n \to f$ pointwise. Hence,
$$ \int_A f \,d\mu = \lim_n \int_A f_n \, d\mu = \lim_n \int f_n \, d\mu = \int f \, d\mu, $$
again by monotone convergence.
In particular, $f$ is integrable (in the usual sense).
Of course, the Amann&Escher integral can not handle the case $\int f \, d\mu = \infty$, because it is an integral that even works with values in Banach spaces. But the above shows that a measurable $f : X \to [0,\infty)$ is integrable (with finite integral) in the usual sense iff it is integrable in the Amann&Escher sense and both integrals coincide.
Ok, here is a proof for the monotonicity of the Amann&Escher integral (also posted as a comment above):
You just have to note that $f \geq 0$ implies $\int_A f \, d\mu \geq 0$ (by linearity).
To see this, note that $|f_+ - g_+| \leq |f-g|$, where $f_+$ is the positive part of $f$, so that if $f_n \to f$ pointwise and $(f_n)_n$ is $L^1$-Cauchy, then $((f_n)_+)_n$ is also $L^1$-Cauchy with $(f_n)_+ \to f$ pointwise. Hence, $\int_A f \, d\mu = \lim_n \int (f_n)_+ \, d\mu \geq 0$.
Shortcut
By Fatou it turns into an amazing one-liner: $$\int|f-s_n|\mathrm{d}\mu\leq\liminf_m\int|s_m-s_n|\mathrm{d}\mu\to0\implies\int|f|\mathrm{d}\mu<\infty$$ Especially one has then: $$\left|\int f\mathrm{d}\mu-\int s_n\mathrm{d}\mu\right|\leq\int|f-s_n|\mathrm{d}\mu\to0\implies\int f\mathrm{d}\mu=\lim_n\int s_n\mathrm{d}\mu$$
Ok, so after some tries I think I got it more or less...
The problem is that being cauchy does not imply having a dominant...
Amann & Escher Integral
(Hard Part)
Suppose $f_0\in\mathcal{L}_\mathfrak{AE}$.
First of all, the Amann & Escher integral is positive: $$f\geq0:\quad0\leq\int|s^+_m-s^+_n|\mathrm{d}\mu\leq\int|s_m-s_n|\mathrm{d}\mu\to0\implies0\leq\lim_n\int s^+_n\mathrm{d}\mu=\int_\mathfrak{AE}f\mathrm{d}\mu$$ (Thanks to the idea by PhoemueX!)
But also the Amann & Escher integral is obviously linear!
So the Amann & Escher integral is monotone: $$f\leq g:\quad0\leq\int_\mathfrak{AE}(f-g)\mathrm{d}\mu=\int_\mathfrak{AE}f\mathrm{d}\mu-\int_\mathfrak{AE}g\mathrm{d}\mu$$
Thus an increasing approximation is monotone and bounded whence cauchy: $$0\leq s_n\uparrow f_0:\quad\int s_{n-1}\mathrm{d}\mu\leq\int s_n\mathrm{d}\mu=\int_\mathfrak{AE}s_n\mathrm{d}\mu\leq\int_\mathfrak{AE}f_0\mathrm{d}\mu$$
Hence the Lebesgue integral agrees with the Amann & Escher integral: $$\int_\mathfrak{L}f_0\mathrm{d}\mu=\lim_n\int s_n\mathrm{d}\mu=\int_\mathfrak{AE}f_0\mathrm{d}\mu<\infty$$
Concluding $f_0\in\mathcal{L}_\mathfrak{L}$.
Lebesgue Integral
(Easy Part)
Conversely, suppose $f_0\in\mathcal{L}_\mathfrak{L}$.
By measurability it admits an increasing approximation so: $$0\leq s_n\uparrow f:\int|s_m-s_n|\mathrm{d}\mu\leq\int(f-s_m)\mathrm{d}\mu+\int(f-s_m)\mathrm{d}\mu\to0$$ (By the way, there's a fresh proof of this by PhoemueX: Lebesgue: Alternative Proof)
Hence the Amann & Escher integral agrees with the Lebesgue integral: $$\int_\mathfrak{AE}f_0\mathrm{d}\mu=\lim_n\int s_n\mathrm{d}\mu=\int_\mathfrak{L}f_0\mathrm{d}\mu$$
Concluding $f_0\in\mathcal{L}_\mathfrak{AE}$.