Exercise 1.3.1 (Basic properties of the simple unsigned integral)

We have the simple representation of $f,g$ in terms of indicator functions

where it is easy to adjust both function to the same number $p=\max <!--\; FUNCTION\; APPLICATION\; -->\{k,n\}$ of indicators. Furthermore, consider a common refinement $A_1,\dots ,A_{k^{\prime }}$ taken from $\mathit{E\#\#F}=\{E_i\cap F_j:1\le i\le k,1\le j\le n\}$. We can then rewrite the simple representation of both $f$ and $g$ using the above set

which will prove useful in what follows.

(i)

(Linearity) $$\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f+g=c_{1}m\left(E_1\right)+\dots +c_{k}m\left(E_1\right)+d_{1}m\left(F_1\right)+\dots +d_{k}m\left(F_k\right)=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f+\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}g$$

(ii)

(Finiteness) We demonstrate both sides of the implication.

• Assume that we have $c_{1}m\left(E_1\right)+\dots +c_{k}m\left(E_1\right)<\infty$. By non-negativity this is equivalent to $\forall <!--\; FUNCTION\; APPLICATION\; -->1\le i\le k:c_{i}m\left(E_i\right)<\infty$. Suppose, for the sake of contradiction that the theorem assertion is false, i.e., one of the following is true

  • $f$ is not finite almost everywhere, i.e., there exists $1\le i\le k$ such that we have a non-null set $m(E_i)\ne 0$ and its associated value $c_i=\infty$. But this leads directly to a contradiction, since $\infty =c_i\cdot m(E_i)\le \mathrm{Simp}{\int }_{{ℝ}^d}f$.
  • We have $m(\mathrm{support}(f))=\infty$. We can represent $\mathrm{support}(f)=E_1\cup \dots \cup E_k$ as a disjoint union $F_1\cup \dots \cup F_n$ from which we conclude by additivity that $m(F_1)+\dots +m(F_n)=m(\mathrm{support}(f))=\infty ⟹\exists <!--\; FUNCTION\; APPLICATION\; -->i:m(F_i)=\infty$. Since $F_i\subseteq \mathrm{support}(f)$ its associated values $d_i$ is non-zero, leading to $\infty =m(F_i)\cdot d_i\le \mathrm{Simp}{\int }_{{ℝ}^d}f$ - a contradiction.

  Since both cases are impossible, the original assertion must be true.

• We have $\mathrm{Simp}{\int }_{{ℝ}^d}f=c_{1}m\left(E_1\right)+\dots +c_{k}m\left(E_1\right)$. We may have two problems: either a $m(E_i)$ is infinite or $c_i$ is infinite. In the former case $m(E_i)=\infty$ we have $E_i⊈\mathrm{support}(f)$ by monotonicity and the assumption (1); thus, $E_i$ cannot have a non-zero $c_i$ associated with it, and we can safely ignore these values as the product $c_{i}m(E_i)=0$ anyway. In the latter case $c_i=\infty$, the associated set $E_i$ must be a null set by assumption (2); thus we can safely ignore the infinite values of $c_i$ as they result in zero.

(iii)

(Vanishing)

• Suppose that $\mathrm{Simp}{\int }_{{ℝ}^d}f=c_{1}m\left(E_1\right)+\dots +c_{k}m\left(E_1\right)=0$. By non-negativity of $c_i$ and $m$ we must have $\forall <!--\; FUNCTION\; APPLICATION\; -->1\le i\le k:c_{i}m(E_i)=0$. There are two possible cases for that, for all $1\le i\le k$ one of the following must hold:

  • $c_i=0$ and $m(E_i)>0$
  • $c_i>0$ and $m(E_i)=0$,
  • or both.

  But this is the definition of almost everywhere $f=0$, as desired.

• Suppose that $f$ is zero everywhere, i.e., whenever $c_i\ne 0$ we have $m(E_i)=0$. Thus, either $m(E_i)>0$ and $c_i=0$ resulting in $c_{i}m(E_i)=0$, or $c_i\ne 0$ and $m(E_i)=0$, resulting in the same quantity. Thus, every $c_{i}m(E_i)=0$ and the whole integral sum must be zero $\mathrm{Simp}{\int }_{{ℝ}^d}f=0$.

(iv)

(Equivalence) On a common refinement we have either (1) $a_i=b_i$ with $m(A_i)\ge 0$ or (2) $a_i\ne b_i$ and $m(A_i)=0$ by theorem assumption. Kicking out all of the zeros $m(A_i)=0$ we observe that the two integral match $\mathrm{Simp}{\int }_{{ℝ}^d}f=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}g$.

(v)

(Monotonicity) Again, on a common refinement $A_1,\dots ,A_{k^{\prime }}$ we kick out all of the values with $m(A_i)=0$; on the rest we have $a_i\le b_i$. By monotonicity of sums, we must have $\mathrm{Simp}{\int }_{{ℝ}^d}f\le \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}g$.

(vi)

(Compatibility with Lebesgue measure) Notice that indicator function has a trivial simple representation $1_E=1\cdot 1_E$, and thus $\mathrm{Simp}{\int }_{{ℝ}^d}{1}_E=1\cdot m(E)=m(E)$ directly by definition.