Exercise 1.4.35 (Basic properties of the unsigned integral)

(i)

Almost everywhere equivalence
Since $f=g$ almost everywhere, then any function $\underline{f}$ minorising $f$ also minorises $g$ almost everywhere. We can easily tweak it on a null set to be a function ${\underline{f}}^{\prime }$ that minorises $g$ everywhere. We then have $\mathrm{Simp}{\int }_X\underline{f}d\mu =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}x{\underline{f}}^{\prime }$ by Exercise 1.4.33. Conversely, any function $\underline{g}$ minorising $g$ will also minorise $f$ $\mu$-almost everywhere, and thus there is a function ${\underline{g}}^{\prime }$ that minorises $f$ everywhere with $\mathrm{Simp}{\int }_X\underline{g}d\mu =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}x{\underline{g}}^{\prime }$. Thus, the set of simple integrals of the minorising functions of both $f$ and $g$ are equal, and so are their supremums.

(ii)

Monotonicity
If almost everywhere $f\le g$, then any simple function $\underline{f}$ minorising $f$ will automatically also minorise $g$ almost everywhere. Changing $\underline{f}$ on a set of null set we get a function ${\underline{f}}^{\prime }$ which minorises $g$ everywhere. In other words, using the insensitivity of the simple integral to changes of the null sets we conclude $$\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_X\underline{f}d\mu :\begin{array}{c}\underline{f}\text{ is simple}\hfill \\ \underline{f}\le f\hfill \end{array}\right\}\subseteq \left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_X\underline{g}d\mu :\begin{array}{c}\underline{g}\text{ is simple}\hfill \\ \underline{g}\le g\hfill \end{array}\right\}$$

Thus, the supremum of the former set is less than the supremum of the latter.

(iii)

Homogeneity
We have to demonstrate that $$c\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_X\underline{f}d\mu :\begin{array}{c}\underline{f}\text{ is simple}\hfill \\ \underline{f}\le f\hfill \end{array}\right\}=\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}x\underline{\mathit{cf}}:\begin{array}{c}\underline{\mathit{cf}}\text{ is simple}\hfill \\ \underline{\mathit{cf}}\le \mathit{cf}\hfill \end{array}\right\}$$

By the properties of supremum this is equivalent to

$$\left\{c\cdot \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_X\underline{f}d\mu :\begin{array}{c}\underline{f}\text{ is simple}\hfill \\ \underline{f}\le f\hfill \end{array}\right\}=\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}x\underline{\mathit{cf}}:\begin{array}{c}\underline{\mathit{cf}}\text{ is simple}\hfill \\ \underline{\mathit{cf}}\le \mathit{cf}\hfill \end{array}\right\}$$

The above is easily verifiable using the analogous property of the simple integrals from Exercise 1.4.33.

(iv)

Superadditivity
We have to verify that $$\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_X\underline{f}d\mu :\begin{array}{c}\underline{f}\text{ is simple}\hfill \\ \underline{f}\le f\hfill \end{array}\right\}+\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_X\underline{g}d\mu :\begin{array}{c}\underline{g}\text{ is simple}\hfill \\ \underline{g}\le g\hfill \end{array}\right\}\le \sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}x\underline{f+g}:\begin{array}{c}\underline{f+g}\text{ is simple}\hfill \\ \underline{f+g}\le f+g\hfill \end{array}\right\}$$

But this follows directly from the fact that for arbitrary simple functions $\underline{f}$ and $\underline{g}$ minorising $f$ and $g$ respectively, the function $\underline{f}+\underline{g}$ minorises $f+g$, and is thus contained in the right-hand side set. Arguing by contradiction we see that the inequality holds.

(v)

Compatibility with the simple integral
We argue by contradiction. The case $$\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_X\underline{f}d\mu :\begin{array}{c}\underline{f}\text{ is simple}\hfill \\ \underline{f}\le f\hfill \end{array}\right\}<\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{X}fd\mu$$

is impossible since the right-hand side term is contained in the left-hand side set. The case

$$\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_X\underline{f}d\mu :\begin{array}{c}\underline{f}\text{ is simple}\hfill \\ \underline{f}\le f\hfill \end{array}\right\}>\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{X}fd\mu$$

is impossible as well, since we would then have a simple functions ${\underline{f}}^{\prime }$ such that $\mathrm{Simp}{\int }_{{ℝ}^d}x{\underline{f}}^{\prime }>\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{X}fd\mu$ despite the fact that ${\underline{f}}^{\prime }\le f$ - a contradiction to monotonicity of the simple integral.

(vi)

Markov’s inequality
We have the trivial pointwise inequality $$\lambda \times 1_{\{x\in X:f(x)\ge \lambda \}}\le f$$

Both sides represent measurable functions; taking integrals we obtain

$$\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\mathit{x\lambda }\times 1_{\{x\in X:f(x)\ge \lambda \}}\le {\int \; <!--\; nolimits\; -->}_{X}fd\mu$$

Using various linearity properties from the previous parts of the exercise we obtain

$$\mu \left(\{x\in X:f(x)\ge \lambda \}\right)\le \frac{1}{\lambda }{\int \; <!--\; nolimits\; -->}_{X}fd\mu .$$

(vii)

(Finiteness)
Suppose for the sake of contradiction that $f$ is not finite almost everywhere, i.e., there exists a measurable set $E\subseteq X$ such that $f(x)=\infty$ for all $x\in E$, and $\mu (E)>0$. For any $K>0$ we then have by Markov’s inequality: $$K\cdot \mu (E)\le K\cdot \mu \left(\{x\in X:f(x)\ge K\}\right)\le {\int \; <!--\; nolimits\; -->}_{X}fd\mu$$

Thus ${\int }_{X}fd\mu$ can get arbitrarily big - a contradiction.

(viii)

Vanishing
My Markov’s inequality we have $$\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}:\mu \left(\left\{x\in X:f(x)\ge \frac{1}{n}\right\}\right)\le n\cdot {\int \; <!--\; nolimits\; -->}_{X}fd\mu =0$$

The sequence of sets $\{x\in X:f(x)\ge 1∕n\}$ is increasing; by monotone convergence from Exercise 1.4.23 we have

$$\mu \left(\left\{x\in X:f(x)\ge 0\right\}\right)=\mu \left(\bigcup _{n=1}^{\infty }\left\{x\in X:f(x)\ge \frac{1}{n}\right\}\right)=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\mu \left(\left\{x\in X:f(x)\ge \frac{1}{n}\right\}\right)=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}0=0$$

(ix)

Horizontal truncation
First, suppose that ${\int }_{X}fd\mu <\infty$; $$\forall <!--\; FUNCTION\; APPLICATION\; -->𝜖>0\exists <!--\; FUNCTION\; APPLICATION\; -->N\in \mathbb{N}\forall <!--\; FUNCTION\; APPLICATION\; -->n>N:{\int \; <!--\; nolimits\; -->}_{X}fd\mu -\min <!--\; FUNCTION\; APPLICATION\; -->\{f,n\}\le 𝜖$$

By supremum definition of integral and by axiom of choice we can find a sequence of simple functions ${\left(h\right)}_{n\in \mathbb{N}}$ minorising $f$ such that

$$\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_X{h}_{n}d\mu ={\int \; <!--\; nolimits\; -->}_{X}fd\mu$$

from below. (If the above quantity ${\int }_{X}fd\mu$ is infinite, then our sequence of simple integrals would be simply unbounded, and the below inequality would prove the unboundedness of $\min <!--\; FUNCTION\; APPLICATION\; -->\{f,n\}$ as well.) For all $n\in \mathbb{N}$ we then have the inequality

$$\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}x\min <!--\; FUNCTION\; APPLICATION\; -->\{h_n,n\}\le \min <!--\; FUNCTION\; APPLICATION\; -->\{f,n\}\le {\int \; <!--\; nolimits\; -->}_{X}fd\mu$$

By definition $\mathrm{Simp}{\int }_X{h}_{n}d\mu ={\sum }_{i=1}^n{a}_i^{(n)}\mu (h_n^{-1}(a_i^{(n)}))$. Since $f$ is finite almost everywhere, so is $h_n$, and so are $a_i^{(n)}<N,N\in \mathbb{N}$ not associated with null sets $h_n^{-1}(a_i^{(n)})$. Thus, for sufficiently large $N\in \mathbb{N}$ we have $\mathrm{Simp}{\int }_{{ℝ}^d}\min <!--\; FUNCTION\; APPLICATION\; -->(h_n,n)=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{h}_n$ for all $n\ge N$. We thus have for all $n\ge N$:

$$\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_X{h}_{n}d\mu \le \min <!--\; FUNCTION\; APPLICATION\; -->\{f,n\}\le {\int \; <!--\; nolimits\; -->}_{X}fd\mu$$

Taking the limits we squeeze the middle term into converging to a desired limit.

(x)

Vertical truncation
By supremum definition of the integral, and by the axiom of choice we can construct a sequence ${(h_m)}_{m\in \mathbb{N}}$ of simple functions minorising $f\cdot 1_{{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n}$ such that for all $m\in \mathbb{N}$ we have $f\cdot 1_{{\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n}-\mathrm{Simp}{\int }_{{ℝ}^d}x{h}_m\le 1∕m$. From monotonicity we have the obvious inequality $$\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}x{h}_m\cdot 1_{E_n}\le f\cdot 1_{E_n}\le f\cdot 1_{\text{⋃}_{n=1}^{\infty }{E}_n}$$

We now take limits as $n\to \infty$. Notice that we have in terms of the simple integral:

$$\begin{array}{llll}\hfill \underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}x(h_m\cdot 1_{E_n})& =\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\sum _{i=1}^k{a}_{m,i}\mu \left({(h_m\cdot 1_{E_n})}^{-1}(a_{m,i})\right)& \hfill & \\ \hfill & =\sum _{i=1}^k{a}_{m,i}\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\mu \left({(h_m\cdot 1_{E_n})}^{-1}(a_{m,i})\right)& \hfill & \end{array}$$

Notice that the sequence of sets ${(h_m\cdot 1_{E_n})}_{n\in \mathbb{N}}$ is increasing in $n$; we can apply the monotone convergence Exercise 1.4.23 to get

$$\begin{array}{llll}\hfill & =\sum _{i=1}^k{a}_{m,i}\cdot \mu \left(\bigcup _{n=1}^{\infty }{(h_m\cdot 1_{E_n})}^{-1}(a_{m,i})\right)& \hfill & \end{array}$$

It can be verified that ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{(h_m\cdot 1_{E_n})}^{-1}(\{c\})={(h_m\cdot 1_{{\bigcup }_{n=1}^{\infty }{E}_n})}^{-1}(\{c\})$ for any $c\in [0,+\infty ]$; thus,

$$\begin{array}{llll}\hfill & =\sum _{i=1}^k{a}_{m,i}\cdot \mu \left({(h_m\cdot 1_{\bigcup _{n=1}^{\infty }{E}_n})}^{-1}(a_{m,i})\right)& \hfill & \\ \hfill & =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}x{h}_m\cdot 1_{\text{⋃}_{n=1}^{\infty }{E}_n}& \hfill & \end{array}$$

Thus, our original inequality reduces to following when taking limits as $n\to \infty$:

$$\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}x{h}_m\cdot 1_{\text{⋃}_{n=1}^{\infty }{E}_n}\le \underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}f\cdot 1_{E_n}\le f\cdot 1_{\text{⋃}_{n=1}^{\infty }{E}_n}$$

Taking limits once again for $m\to \infty$ yields the desired results.

(xi)

Restriction
A simple function $h_n$ minorises $f{1}_Y$ if and only if $h_n{\text{↾}}_Y$ minorises $f{\text{↾}}_Y$. But $h_n=h_n{1}_Y=h_n{\text{↾}}_Y$ on their supports. Taking simple integrals we see that $\mathrm{Simp}{\int }_X{h}_{n}d\mu =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}x{h}_n{1}_Y=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}x{h}_n{\text{↾}}_Y{\text{↾}}_Y$. Thus, we take supremums over the same sets of simple integrals, and the claim follows.

Let $(X,B,\mu )$ be a measure space and let $f:X\to [0,+\infty ]$ be measurable. If $E_1\subseteq E_2\subseteq \cdots$ is an increasing sequence of $B$-measurable sets, then

First, assume $f$ is simple. Then $f={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^k{a}_i{1}_{A_i}$, a finite combination of indicators of measurable sets. Let $E={\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n$. Fix a natural number $m$. Then

Similarly for $E$. This implies:

Taking $m\to \infty$ in the above identity and using continuity from below (upward monotone convergence), we get that

Re-inserting the above equation gives:

By the definition of $E$, this simplifies to:

This proves the claim for the case when $f$ is simple. Now let $f$ be measurable generally. Abbreviate the sets the integrals are defined on by:

We have to prove $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}(\sup <!--\; FUNCTION\; APPLICATION\; -->A_n)=\sup <!--\; FUNCTION\; APPLICATION\; -->A$. Note that for each $n$, any $f\cdot 1_{E_n}\le f\cdot 1_E$, since for $x\in E_n$, we have $x\in E$, and for $x\notin E_n$, $(f\cdot 1_{E_n})(x)=0\le (f\cdot 1_E)(x)$. This implies that any function $g_n\le f\cdot 1_{E_n}\le f\cdot 1_E$, so:

Hence $\sup <!--\; FUNCTION\; APPLICATION\; -->A_n\le \sup <!--\; FUNCTION\; APPLICATION\; -->A$ for all $n$ and

Now assume, to the contrary, $\sup <!--\; FUNCTION\; APPLICATION\; -->A>\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\sup <!--\; FUNCTION\; APPLICATION\; -->A_n$. Similarly to the argument above, note that $f\cdot 1_{E_n}\le f\cdot 1_{E_{n+1}}$, so any function $g_n\le f\cdot 1_{E_n}\le f\cdot 1_{E_{n+1}}$, further implying:

Then ${\{\sup <!--\; FUNCTION\; APPLICATION\; -->A_n\}}_{n=1}^{\infty }$ is an increasing sequence.

Since $\sup <!--\; FUNCTION\; APPLICATION\; -->A>\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\sup <!--\; FUNCTION\; APPLICATION\; -->A_n$, there is a simple function $h\le f\cdot 1_E$ such that:

Given $h$, create the sequence of simple functions ${\{g_n\}}_{n=1}^{\infty }$ with $g_n=h\cdot 1_{E_n}$. Then:

Since $h$ is zero outside of $E$, we have:

On the other hand, for each $g_n=h\cdot 1_{E_n}\le (f\cdot 1_E)\cdot 1_{E_n}=f\cdot 1_{E_n}$, so $\mathrm{Simp}<!--\; FUNCTION\; APPLICATION\; -->{\mathrm{\int \; <!--\; nolimits\; -->}<!--\; FUNCTION\; APPLICATION\; -->}_X{g}_n\mathit{d\mu }\in A_n$. Then:

a contradiction to the observation above. Hence $\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\sup <!--\; FUNCTION\; APPLICATION\; -->A_n=\sup <!--\; FUNCTION\; APPLICATION\; -->A$ and the claim follows by the definitions of the integral.