General definition of the integral of a nonnegative function and proving two theorems

Question

The Problem

I am really struggling to wrap my head around the definition of the integral in measure theory. According to Axler, the integral of a nonnegative function is defined as follows:

Suppose $(X, \mathcal S,\mu)$ is a measure space and $f : X \to [0,\infty]$ a measurable function. The integral of $f$ with respect to $\mu$, denoted $\int f\, \mathrm d \mu$ is defined by \begin{multline} \int f\,\mathrm d\mu = \sup \bigg\{\sum_{k=1}^{m} a_k \mu(A_k) : \sum_{k=1}^{m} a_k\chi_{A_k} \leq f,\quad m\in\mathbb Z^+,\quad a_1,\ldots,a_m > 0\quad \text{and} \quad A_1,\ldots, A_m\in\mathcal S\bigg\}. \end{multline}

I've been asked to prove Theorems 3.10 and 3.12 in the linked book, that the integral is order-preserving, as in if $f \leq g$, then \begin{equation} \int f\,\mathrm d \mu \leq \int g \,\mathrm d \mu \end{equation} and that the integral is positively homogeneous, as in if $c\in\mathbb R : c \geq 0$, then \begin{equation} \int cf\,\mathrm d\mu = c\int f\,\mathrm d\mu\,. \end{equation} The book gives hints in both cases, namely that

3.10

The supremum defining $\int f\,\mathrm d\mu$ is taken over a subset of the corresponding set for the supremum defining $\int g\,\mathrm d\mu$.

and in the case of positive homogeneity:

3.12

The supremum defining $\int cf\,\mathrm d\mu$ is taken over a set consisting of $c$ times the set, whose supremum defines $\int f\,\mathrm d\mu$.

You would think that this was enough of a hint, but I am having trouble justifying the steps in the "calculations".

3.10

Let $f,g : X\to[0,\infty]$ be measurable, meaning that \begin{equation} F = \bigcup_{k=1}^{m} F_k = f^{-1}([0,\infty]) \in\mathcal S \quad \text{and}\quad G = \bigcup_{k=1}^{m} G_k = g^{-1}([0,\infty]) \in\mathcal S\,. \end{equation} If we now take the hint, choose the sets $F\subseteq G \subseteq X$ so that each of the sums in the definition of the integral reach their supremums. Then \begin{align} \int_F f\,\mathrm d\mu &= \sup\bigg\{ \sum_{k=1}^{m} a_k \mu(F_k) : \sum_{k=1}^{m} a_k \chi_{F_k} \leq f \bigg\}\\ &\leq \sup\bigg\{ \sum_{k=1}^{m} a_k \mu(G_k) : \sum_{k=1}^{m} a_k \chi_{G_k} \leq g \bigg\}\\ &= \int_G g\,\mathrm d\mu, \end{align} where the inequality follows from the fact that measure preserves order, as in if $F \subseteq G$, then $\mu(F) \leq \mu(G)$. This seems to be enough of a proof in my head, but I am not totally convinced. As for the next theorem...

3.12

Let $f$ be a measurable function and $c \geq 0$, and once again choose the pre-image of $f$, $F$, so that the sum in the definition of the integral reaches its supremum and also define the set \begin{equation} cF = \{ca : a\in F\} = c\bigcup_{k=1}^{m} F_k = \bigcup_{k=1}^{m} cF_k\,. \end{equation} Then \begin{align} \int_{cF} cf\,\mathrm d\mu &= \sup\bigg\{ \sum_{k=1}^{m} a_k \mu(cF_k) : \sum_{k=1}^{m} a_k \chi_{cF_k} \leq cf \bigg\}\\ &= \sup\bigg\{ \sum_{k=1}^{m} a_k c\mu(F_k) : \sum_{k=1}^{m} a_k \chi_{cF_k} \leq cf \bigg\}\\ &= \sup\bigg\{c \sum_{k=1}^{m} a_k \mu(F_k) : \sum_{k=1}^{m} a_k \chi_{cF_k} \leq cf \bigg\}\\ &= c\sup\bigg\{\sum_{k=1}^{m} a_k \mu(F_k) : \sum_{k=1}^{m} a_k \chi_{cF_k} \leq cf \bigg\}\\ &= c\int_F f\,\mathrm d\mu\,. \end{align} This is where I'm not getting it. The proof works if \begin{equation} \mu(cF_k) = c\mu(F_k), \end{equation} but I don't remember such a result from the book or our exercise sessions. What I'm especially interested in is, is my understanding of the idea behind the proof correct, or have I totally just missed the target?

Answer 1

There are some correct ideas in your proofs, but I think you may be over-complicating things a bit. I would consider the following a complete proof for 3.10:

Proof. If $\{F_k\}_{k=1}^m\subset\mathcal{S}$ is a collection of subsets and $\{a_k\}_{k=1}^m$ is a set of positive real numbers such that $\sum_{k=1}^ma_k\chi_{F_k}\leq f$, then $f\leq g$ implies $\sum_{k=1}^ma_k\chi_{F_k}\leq g$ as well. It follows that every element in the set $$\left\{\sum_{k=1}^ma_k\mu(F_k):\sum_{k=1}^ma_k\chi_{F_k}\leq f\right\}$$ also belongs to the set $$\left\{\sum_{k=1}^mb_k\mu(G_k):\sum_{k=1}^mb_k\chi_{G_k}\leq g\right\},$$ or simply $$\left\{\sum_{k=1}^ma_k\mu(F_k):\sum_{k=1}^ma_k\chi_{F_k}\leq f\right\}\subset\left\{\sum_{k=1}^mb_k\mu(G_k):\sum_{k=1}^mb_k\chi_{G_k}\leq g\right\}.$$ This containment (hinted at in the book) implies

\begin{align} \int f\,d\mu&=\sup\left\{\sum_{k=1}^ma_k\mu(F_k):\sum_{k=1}^ma_k\chi_{F_k}\leq f\right\}\\ &\leq\sup\left\{\sum_{k=1}^mb_k\mu(G_k):\sum_{k=1}^mb_k\chi_{G_k}\leq g\right\}=\int g\,d\mu.\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\square \end{align}

Note that it was not necessary to explicitly deal with the measurability of $f$ and $g$, nor was it important to mention domains of integration. It is implied in the statements "$\{F_k\}_{k=1}^m\subset\mathcal{S}$", "$\sum_{k=1}^ma_k\chi_{F_k}\leq f$", and "$f\leq g$" that $f$ and $g$ are both defined on all $F_k$.

It is true that $F=\bigcup_{k=1}^mF_k\subset f^{-1}[0,\infty]$, but the requirement that $a_k>0$ for all $k$ means you cannot assume $\bigcup_{k=1}^mF_k=f^{-1}[0,\infty]$ (For example, what would this mean if $f$ were the zero function on $X=[0,1]$?). You also cannot assume "the sums reach their suprema". For example, the integral of $f(x)=1/x$ over $X=(0,1]$ (given by the supremum definition) is $\infty$, but you will never find a finite collection of Lebesgue measurable $F_k\subset(0,1]$ and $a_k>0$ such that $\sum a_k\mu(F_k)=\infty$.

Looking back at the proof of 3.10, notice that all that really needed to be shown was $$\left\{\sum_{k=1}^ma_k\mu(F_k):\sum_{k=1}^ma_k\chi_{F_k}\leq f\right\}\subset\left\{\sum_{k=1}^mb_k\mu(G_k):\sum_{k=1}^mb_k\chi_{G_k}\leq g\right\},$$ then the proof followed through suprema. Try to handle 3.12 similarly: Start by showing $$\left\{\sum_{k=1}^ma_k\mu(F_k):\sum_{k=1}^ma_k\chi_{F_k}\leq cf\right\}=\left\{c\sum_{k=1}^mb_k\mu(F_k):\sum_{k=1}^mb_k\chi_{F_k}\leq f\right\},$$ then take suprema. One more important thing to look out for: In general, $X$ does not have to be a subset of the complex numbers, so it often doesn't make sense to multiply sets $F\subset X$ by complex numbers $c$. This means the statement "$\mu(cF)=c\mu(F)$" is false in general, so don't use it. Showing that the two sets above are equal does not require passing constants in and out of measures.

Answer 2

All this has nothing to do with $\mu(cF)$. If $F$ is a subset of a general measure space there's no such thing as $cF$.

If $\phi$ is simple, $0\le\phi\le f$ and $c\ge0$ then $c\phi$ is simple and $0\le c\phi\le cf$. conversely, if $\psi$ is simple and $0\le\psi\le cf$ then $\psi=c\phi$ where $0\le\phi\le f$.

So: The set of simple functions that arises in the definition of $\int cf$ is exactly $c$ times the set of simple functions that arises in the definition of $\int f$. If $A$ is the set of simple functions $\phi$ with $0\le\phi\le f$ and $B$ is the set of simple functions $\phi$ with $0\le\phi\le cf$ then $B=cA$. Hence similarly for the sets of integrals of simple functions such that etc.