Exercise 1.3.10 (Basic properties of lower Lebesgue integral)

Some of the assertions will follow directly from their counterparts for simple integrals from Exercise 1.3.1.

(i)

Compatibility with the simple integral
This assertion is rather trivial since for any simple function $g\le f$ we have $\mathrm{Simp}{\int }_{{ℝ}^d}g\le \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f$; thus, $\mathrm{Simp}{\int }_{{ℝ}^d}f$ is the supremum over all of the integrals minorizing $f$. Similarly with the upper integral.

(ii)

Monotonicitysimple function that minorizes $f$ almost everywhere will also minorize $g$ almost everywhere by the property $f\stackrel{\mathit{ae}}{\text{≤}}g$. Thus the set of all simple integrals minorizing $f$ is a subset of the set of all simple integrals minorizing $g$; the supremum of the latter is hence larger than the supremum of the former. A symmetric argument gives us the inequality for the upper integral.

(iii)

Homogeneity
Let $c\in [0,+\infty )$. We have to demonstrate that $$\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}g:\begin{array}{c}g\text{ is simple}\hfill \\ 0\le g\le \mathit{cf}\hfill \end{array}\right\}=c\times \sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}h:\begin{array}{c}h\text{ is simple}\hfill \\ 0\le h\le f\hfill \end{array}\right\}$$

By the homogeneity of the $\sup <!--\; FUNCTION\; APPLICATION\; -->$ and $\inf <!--\; FUNCTION\; APPLICATION\; -->$ this is the same as

$$\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}g:\begin{array}{c}g\text{ is simple}\hfill \\ 0\le g\le \mathit{cf}\hfill \end{array}\right\}=\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{c\times \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}h:\begin{array}{c}h\text{ is simple}\hfill \\ 0\le h\le f\hfill \end{array}\right\}$$

But this is easy to prove using the analogous property of simple integrals from Exercise 1.3.1.

• Pick an arbitrary simple function $g$ minorizing $\mathit{cf}$. Obviously, $1∕c\cdot g$ minorizes $f$. Thus, $g∕c$ is contained in the latter set, and $\mathrm{Simp}{\int }_{{ℝ}^d}g=c\times \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}g∕c$ by the properties of the simple integral.
• Similarly, pick an arbitrary simple $h$ minorizing $f$, and consider $c\times \mathrm{Simp}{\int }_{{ℝ}^d}h=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\mathit{ch}$. Since $\mathit{ch}$ minorizes $\mathit{cf}$, the integral $\mathrm{Simp}{\int }_{{ℝ}^d}\mathit{ch}$ must be contained in the former set.

(iv)

Equivalence
Let $N\subseteq {ℝ}^d$ be the null set on which $f\ne g$. We want to demonstrate that $$\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\underline{f}:\begin{array}{c}\underline{f}\text{ is simple}\hfill \\ 0\le \underline{f}\le f\hfill \end{array}\right\}=\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\underline{g}:\begin{array}{c}\underline{g}\text{ is simple}\hfill \\ 0\le \underline{g}\le g\hfill \end{array}\right\}$$

It is easy to demonstrate that both sets in question are equal. Pick an arbitrary simple lower cover $\underline{f}$ of $f$. Since $\underline{f}\le f$ and $f\stackrel{\mathit{ae}}{\text{=}}g$, we must have $\underline{f}\stackrel{\mathit{ae}}{\text{≤}}g$ (since any $x\in {ℝ}^d$ such that $\underline{f}(x)>g(x)$ also satisfies $g(x)\ne f(x)$ and thus $x\in N$). Set $\underline{g}(x)=\underline{f}(x)$ on ${ℝ}^d∕N$ and $\underline{f}(x)=0$ on the Lebesgue measurable set $N$. Then, $\underline{g}$ is a simple function minorizing $g$, but we also have by noise-tolerance of simple integrals $\mathrm{Simp}{\int }_{{ℝ}^d}g=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f$ contained in the set of simple integrals minorizing $f$. A analogous line of argumentation demonstrates the converse statement.
The proof for the upper Lebesgue integral is analogous.

(v)

Superadditivity of the lower integralhave to demonstrate that $$\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\underline{h}:\begin{array}{c}\underline{h}\text{ is simple}\hfill \\ 0\le \underline{h}\le f+g\hfill \end{array}\right\}\ge \sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\underline{f}:\begin{array}{c}\underline{f}\text{ is simple}\hfill \\ 0\le \underline{f}\le f\hfill \end{array}\right\}+\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\underline{g}:\begin{array}{c}\underline{g}\text{ is simple}\hfill \\ 0\le \underline{g}\le g\hfill \end{array}\right\}$$

Recall that $\sup <!--\; FUNCTION\; APPLICATION\; -->(A+B)=\sup <!--\; FUNCTION\; APPLICATION\; -->(A)+\sup <!--\; FUNCTION\; APPLICATION\; -->(B)$, where $A$ and $B$ denote the right-hand side sets in the above inequality. If we manage to show that the left hand side set contains $A+B$, then the assertion follows. Pick some simple lower cover $\underline{f}$ of $f$ (from $A$) and $\underline{g}$ of $g$ (from $B$). Then, $\underline{f}+\underline{g}$ is a simple lower cover of $f+g$, and is thus contained in the left-hand side set. By monotonicity of the supremum the inequality is true.

(vi)

Subadditivity of the upper integralhave to demonstrate that $$\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\overline{h}:\begin{array}{c}\overline{h}\text{ is simple}\hfill \\ f+g\le \overline{h}\hfill \end{array}\right\}\le \inf <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\overline{f}:\begin{array}{c}\overline{f}\text{ is simple}\hfill \\ f\le \overline{f}\hfill \end{array}\right\}+\inf <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\overline{g}:\begin{array}{c}\overline{g}\text{ is simple}\hfill \\ g\le \overline{g}\hfill \end{array}\right\}$$

Recall that $\inf <!--\; FUNCTION\; APPLICATION\; -->(A+B)=\inf <!--\; FUNCTION\; APPLICATION\; -->(A)+\inf <!--\; FUNCTION\; APPLICATION\; -->(B)$, where $A$ and $B$ denote the right-hand side sets in the above inequality. If we manage to show that the left hand side set contains $A+B$, then the assertion follows. Pick an arbitrary simple upper cover $\overline{f}$ of $f$ and $\overline{g}$ of $g$. Then, $\overline{f}+\overline{g}$ is a simple upper cover of $f+g$, and is thus contained in the left-hand side set. By monotonicity of the infimum the inequality is true.

(vii)

Divisibilityhave to demonstrate that for a Lebesgue measurable set $E\subseteq {ℝ}^d$: $$\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}h:\begin{array}{c}h\text{ is simple}\hfill \\ 0\le h\le f\hfill \end{array}\right\}=\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{h}_1:\begin{array}{c}{h}_1\text{ is simple}\hfill \\ 0\le h_1\le f\cdot 1_E\hfill \end{array}\right\}+\sup <!--\; FUNCTION\; APPLICATION\; -->\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}g:\begin{array}{c}{h}_2\text{ is simple}\hfill \\ 0\le h_2\le f\cdot 1_{{ℝ}^d∕E}\hfill \end{array}\right\}$$

Recall that $\sup <!--\; FUNCTION\; APPLICATION\; -->(A+B)=\sup <!--\; FUNCTION\; APPLICATION\; -->(A)+\sup <!--\; FUNCTION\; APPLICATION\; -->(B)$. Thus if we manage to demonstrate that

$$\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}h:\begin{array}{c}h\text{ is simple}\hfill \\ 0\le h\le f\hfill \end{array}\right\}=\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{h}_1:\begin{array}{c}{h}_1\text{ is simple}\hfill \\ 0\le h_1\le f\cdot 1_E\hfill \end{array}\right\}+\left\{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}g:\begin{array}{c}{h}_2\text{ is simple}\hfill \\ 0\le h_2\le f\cdot 1_{{ℝ}^d∕E}\hfill \end{array}\right\}$$

then we are done. Denote the left-hand side set by $C$, and the two sets on the right-hand side by $A$ and $B$.

• Pick an arbitrary $\mathrm{Simp}{\int }_{{ℝ}^d}h\in C$ of the lower cover $h$ of $f$. Then $h\cdot 1_E$ is still simple (if $I_1,\dots ,I_n\subset {ℝ}^d$ with the values $c_1,\dots ,c_n$ is a simple partition of $f$, then $I_1\cap E,\dots ,I_n\subset {ℝ}^d$ is a simple partition of $f\cdot 1_E$ associated with values $c_1,\dots ,c_n$). Furthermore $h\cdot 1_E\le f\cdot 1_E$ on $E$. Thus we have found a simple lower cover of $f\cdot 1_E$, and we have $\mathrm{Simp}{\int }_{{ℝ}^d}h\cdot 1_E\in A$. Similarly, $h\cdot 1_{{ℝ}^d∕E}$ constitutes for a simple lower cover of $f\cdot 1_{{ℝ}^d∕E}$, and we obtain $\mathrm{Simp}{\int }_{{ℝ}^d}h\cdot 1_E\in B$. By divisibility of the simple integral we have $\mathrm{Simp}{\int }_{{ℝ}^d}h=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}h\cdot 1_E+\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}h\cdot 1_{{ℝ}^d∕E}$, and we are done.
• Pick an arbitrary $\mathrm{Simp}{\int }_{{ℝ}^d}{h}_1\in A$ for some lower cover $h_1$ of $f\cdot 1_E$ and an arbitrary $\mathrm{Simp}{\int }_{{ℝ}^d}{h}_2\in B$ for some lower cover $h_2$ of $f\cdot 1_{{ℝ}^d∕E}$. It is easy to verify that $h_1+h_2$ then is a simple functions which is an inner cover of $f$, and we thus have $\mathrm{Simp}{\int }_{{ℝ}^d}{h}_1+\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{h}_2=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{h}_1+h_2\in C$.

(viii)

Horizontal truncationconsider the case where $f$ might not be finite almost everywhere. Denote the domain of $f$ with such property by $B:=\{x\in {ℝ}^d:f(x)=\infty \}$. We then have $\underline{{\int }_{{ℝ}^d}}f\ge \underline{{\int }_{{ℝ}^d}}f\cdot 1_B=\underline{{\int }_{{ℝ}^d}}\infty \cdot 1_B=\mathrm{Simp}{\int }_{{ℝ}^d}\infty \cdot 1_B=\infty$ by the part (i) of this Exercise. Our the sequence is also unbounded: $$\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}:\underline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}\min <!--\; FUNCTION\; APPLICATION\; -->\left\{f,n\right\}\ge \underline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}\min <!--\; FUNCTION\; APPLICATION\; -->\left\{f,n\right\}\cdot 1_B=\underline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}n\cdot 1_B=n\cdot 1_B$$

Thus it will become $\infty$ eventually. Consider a less degenerate case where $f$ is assumed to be finite almost everywhere (which, when working with measure and integrals, is equivalent to being finite everywhere). We want to demonstrate

$$\forall <!--\; FUNCTION\; APPLICATION\; -->𝜖>0\exists <!--\; FUNCTION\; APPLICATION\; -->N\in \mathbb{N}\forall <!--\; FUNCTION\; APPLICATION\; -->n\ge N:\underline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}f-\underline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}\min <!--\; FUNCTION\; APPLICATION\; -->\left\{f,n\right\}\le 𝜖$$

By the definition of the lower Lebesgue integral we can find a (by almost everywhere finite assumption bounded) simple function $h_n$ such that $\underline{{\int }_{{ℝ}^d}}f-\mathrm{Simp}{\int }_{{ℝ}^d}{h}_n\le 1∕n$. Furthermore, for each $h_n$ we can find an integer $N_n\in \mathbb{N}$ such that $h_n<N_n$ by the boundedness property of our simple function. Combining this with $\min <!--\; FUNCTION\; APPLICATION\; -->\left\{f,n\right\}\le f$ we see the sandwich inequality

$$\underline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}{h}_n\le \underline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}\min <!--\; FUNCTION\; APPLICATION\; -->\left\{f,N_n\right\}\le \underline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}f$$

Taking limits as $n\to \infty$ we obtain the desired property.

(ix)

Vertical truncationuse a strategy similar to those used in the proof of the horizontal trunction. Pick an arbitrary $𝜖>0$ and pick a simple function $h_{𝜖}$ such that $\underline{{\int }_{{ℝ}^d}}f-\mathrm{Simp}{\int }_{{ℝ}^d}{h}_{𝜖}\le 𝜖$. Let $(c_1,I_1),\dots ,(c_k,I_k)$ be the simple representation of $h_{𝜖}$. Notice that $h_{𝜖}\cdot 1_{\{x:\parallel x\parallel \le n\}}$ is also a simple function with a simple representation $(c_1,I_1\cap \{\parallel x\parallel \le n\}),\dots ,(c_k,I_k\cap \{\parallel x\parallel \le n\})$, and it minorizes $f\cdot 1_{\{x:\parallel x\parallel \}}$. We have a sandwich inequality $$\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{h}_{𝜖}\cdot 1_{\{x:\parallel x\parallel \le n\}}\le \underline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}f\cdot 1_{\{x:\parallel x\parallel \le n\}}\le \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f$$

We now take the limits from both sides. Notice that we can employ Exercise 1.2.11 to compute

$$\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}h\cdot 1_{\{x:\parallel x\parallel \le n\}}=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{c}_1\cdot m\left(I_1\cap \{\parallel x\parallel \le n\}\right)+\dots <!--\; FUNCTION\; APPLICATION\; -->+c_k\cdot m\left(I_k\cap \{\parallel x\parallel \le n\}\right)=c_{1}m(I_1)+\dots <!--\; FUNCTION\; APPLICATION\; -->+c_{k}m(I_k)$$

Thus,

$$\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{h}_{𝜖}\le \underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\underline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}f\cdot 1_{\{x:\parallel x\parallel \le n\}}\le \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f$$

Taking the second limit $𝜖\to 0$ we squeeze the middle term into

$$\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}\underline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}f\cdot 1_{\{x:\parallel x\parallel \le n\}}=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f$$

(x)

Reflection

• Pick an arbitrary $0\le f^{\prime }\le f$, $f^{\prime }$ is a simple function. Since $f+g$ is simple, we observe that $g^{\prime }:=f+g-f^{\prime }$ is also a simple function such that $g^{\prime }\ge g$. Obviously we have

  $$\begin{array}{llll}\hfill & \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f+g=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}^{\prime }+g^{\prime }& \hfill & \\ \hfill & =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}^{\prime }+\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{g}^{\prime }\ge \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}^{\prime }+\overline{{\int <!--\; nolimits\; -->}_{{ℝ}^d}}g& \hfill & \end{array}$$

  Taking the supremum for $0\le f^{\prime }\le f$ we obtain

  $$\begin{array}{lll}\hfill \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f+g\ge \underline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}f+\overline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}g& & \hfill \end{array}$$

• Similarly we have

  $$\begin{array}{llll}\hfill \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f+g=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}^{\prime }+g^{\prime }& =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}^{\prime }+\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{g}^{\prime }\le \underline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}f+\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{g}^{\prime }& \hfill & \end{array}$$

  Taking infimums over $g^{\prime }\ge g$ we obtain

  $$\begin{array}{llll}\hfill \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f+g=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}^{\prime }+g^{\prime }& =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}^{\prime }+\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{g}^{\prime }\le \underline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}f+\overline{{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}}g& \hfill & \end{array}$$