# Amann & Escher Integral vs. Lebesgue Integral

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.)

• For simple functions this follows from the definition. I am $100%$ sure that Amann & Escher prove the monotone convergence theorem for their integral (which btw. is nothing but the Bochner integral). This will allow you to conclude equality for $f \geq 0$ measurable (at least if $f$ is integrable). One more way to see this (this is more or less the proof of monotone convergence): If $f_n$ increases to $f$, it is easy to see that the sequence is automatically $L^1$ Cauchy if the sequence if integrals is bounded (i.e.. if $f$ is integrable). – PhoemueX Dec 4 '14 at 7:58
• @PhoemueX: For simple functions they just take the direct definition via their sum. Fine. Unfortunately, they don't prove monotone convergence for the Bochner integral: They stay within the Banach space case and I couldn't find it browsing through the book. After this they jump to positive functions and start all over again with the Lebesgue integral, prove monotone convergence for it and mention in a remark without a proof that the so obtained notion of integral coincides with the earlier one. 'dough' – C-Star-W-Star Dec 4 '14 at 12:29
• @PhoemueX: Ok, for an increasing approximation it follows if one can check integrability but w.r.t. the Lebesgue integral. Now that is the problem as this is precisely what one would like to check. So the whole thing turning it around again and again becomes a circular argument. I really don't know how to solve this dilemma. =/ – C-Star-W-Star Dec 4 '14 at 12:35

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 $f_0\in\mathcal{L}_\mathfrak{AE}$.

• Ok, you did beat me 1 minute :) (and in brevity). – PhoemueX Dec 4 '14 at 13:23
• Haha ^^ ok but mine still lacks at the first point ;). Let me check yours. Can you salvage my first point? – C-Star-W-Star Dec 4 '14 at 13:25
• You just have to note that $f \geq 0$ implies $\int_A f \, d\mu \geq 0$ (by linearity). To see this, note that $|f_+ - g_+| \leq |f-g|$, 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, $\int_A f \, d\mu = \lim_n \int (f_n)_+ \, d\mu \geq 0$. – PhoemueX Dec 4 '14 at 13:33
• Hmmm... let me think. – C-Star-W-Star Dec 4 '14 at 13:37
• @PhoemueX: Ok so everything is fine noting that $|f_+-g_+|\leq|f-g|$. Great!! =D – C-Star-W-Star Dec 4 '14 at 13:49