I have the following statements in what I'm reading:
Let $(X,\mathcal{F},\mu)$ be a measure space, and let
$$\mathcal{M}(X,\mathcal{F})^{+}=\{f:X\to\mathbb{R}:\text{f is }\mathcal{F}\text{-measureable and }f(x)\geq 0\forall x\in X\}$$
and
$$\Phi_f^+=\{\rho\in\mathcal{M}(X,\mathcal{F})^{+}:\rho\text{ is simple and }0\leq\rho(x)\leq f(x)\forall x\in X\}$$
If $f\in\mathcal{M}(X,\mathcal{F})^{+}$ we define the integral of $f$ with respect to $\mu$ as:
$$\int f d\mu=\sup_{\rho \in \Phi^{+}_{f}}\int\rho d\mu$$
I don't really understand the last statement, so let me explain what I do understand: I have that when $\rho$ is a simple function, I have:
$$\int\rho d\mu=\sum_{i=1}^{n}a_i\mu(E_i)$$
This bascially says, if we have functions that creates some squares, we can easily get the area under the graph by adding the height $a_i$ $n$-times, as $\rho^{-1}(a_i)=E_i\quad\iff\quad \rho(E_i)=a_i$. Furthermore, as it's the integral with respect to $\mu$ we then for every different value $a_i$, which is the $E_i$ in the range of the function, multiply with some function $\mu$ value for $E_i$ that is $\mu(E_i)$.
This the whole idea with this question, namely what is meant with:
$$\int f d\mu=\sup_{\rho \in \Phi^{+}_{f}}\int\rho d\mu$$
So we say the integral of $f$ with respect $\mu$ is:
Given that $f$ is a positive function with the codomain $\mathbb{R}$ (and it's measurable), then this is defined as being: The supremum to the integral of the simple function $\rho$ (I have defined $\rho$ earlier), given that $\rho\in\Phi_f^+$ (said otherwise, that $\rho$ has to be in the codomain of $f$, is simple and is less than or equal to $f$, where the codomain of $f$ is $\mathcal{M}(X,\mathcal{F})^{+}$).
What does it mean to find a supremum to $\int \rho d\mu$ given $\rho\in\Phi_f^+?$
Or asked as I understand it: What does it mean to find the supremum to $\sum_{i=1}^{n}a_i\mu(E_i)$ for $\rho$ in the comdomain of $f$ (where I with comdomain of $f$ mean $\mathcal{M}(X,\mathcal{F})^{+}$), with $\rho$ being simple and less than $f$, namely that $\rho\in\Phi_f^+$?
My thoughts:
In the proof of the Riemann integral we see that infimum and supremum goes towards the same limit in the sum when we make the interval on the x-axis smaller and smaller, that we fx. call $I$, that is the Rieman integral. When we add heights and lengths in $\sum_{i=1}^{n}a_i\mu(E_i)$ and then multiply each rectangle with some function $\mu$ to every rectangle $\mu(E_i)$, namely $\mu(E_i)$ – how can we find a supremum? Why do we need supremum? In the Riemann integral we get different values for $I$ with different intervals on the x-axis, so it makes sense for me to take the supremum, but in this case, I don't see the use of supremum?
I really can't get my head around this. If someone, really slowly, can explain this to me, I would greatly appreciate it.
Best Answer
Fortunately, the definition is really not saying very much. For each simple function $0\le \rho\le f$, we know how to compute the integral $\int \rho\,d\mu$. So, consider the set $\mathcal I$ of all such integrals (note these are just numbers, each representing the "area under the graph" of some $\rho$): $$ \mathcal I = \left\{ \int \rho\,d\mu : 0\le \rho\le f,\ \text{$\rho$ is simple}\right\} = \left\{ \int \rho\,d\mu : \rho\in\Phi_f^+\right\}. $$ Now, if $f\ge 0$ is measurable, we define $$ \int f\,d\mu \stackrel{\text{def}}{=} \sup \mathcal I. $$ Sometimes $\sup\mathcal I$ is written in the notation $$ \int f\,d\mu \stackrel{\text{def}}{=} \sup_{\rho\in\Phi_f^+}\int \rho\,d\mu. $$ The upshot is that if you know how to take the supremum of a set of real numbers, then you know what $\int f\,d\mu$ is.