Exercise 1.3.2 (Basic properties of complex simple integral)

We have thus introduced three nested notions of integrals

• simple integral of non-negative functions $f:{ℝ}^d\to [0,+\infty ]$
• simple integrals of real valued functions $f:{ℝ}^d\to {ℝ}^{\text{∗}}$
• simple integrals of complex valued functions $f:{ℝ}^d\to ℂ$

Our strategy is to prove the theorem assertion for the integrals in the above sequence, each time generalising the result to a more general notion of the integral. We have already proven these assertions for the simple integrals of non-negative functions in [?]. We now prove it for real valued functions in the first part of the proof, and only then for complex valued functions in the second part.

Real-valued simple integral. As agreed, first assume that $f,g:{ℝ}^d\to {ℝ}^{\text{∗}}$ are real-valued.

• (Linearity)
  Imagine for a second that we have proven the following identity:

  $$\begin{array}{llll}\hfill & {(f+g)}_{\text{+}}-{(f+g)}_{\text{−}}=(f_{\text{+}}-f_{\text{−}})+(g_{\text{+}}-g_{\text{−}})& \hfill & \end{array}$$

  Using this, we would have managed to represent the identities only using the operation of addition $\text{+}$ as follows (this is necessary to make use of the definition of the signed integral).

  $${(f+g)}_{\text{+}}+f_{\text{−}}+g_{\text{−}}={(f+g)}_{\text{−}}+f_{\text{+}}+g_{\text{+}}$$

  In other words, both sides would then be represented by unsigned simple functions, and so, using the linearity of the unsigned integral from [?] we would be able to arrive at the following step:

  $$\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{(f+g)}_{\text{+}}+\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}_{\text{−}}+\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{g}_{\text{−}}=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{(f+g)}_{\text{−}}+\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}_{\text{+}}+\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{g}_{\text{+}}.$$

  Simple arithmetic manipulations would then lead us to

  $$\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{(f+g)}_{\text{+}}-\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{(f+g)}_{\text{−}}=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}_{\text{+}}-\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}_{\text{−}}+\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{g}_{\text{+}}-\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{g}_{\text{−}}.$$

  And thus,

  $$\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}(f+g)=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f+\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}g$$

  as desired.
  It is thus only left to verify that the identity we have used

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->x\in {ℝ}^d:{(f+g)}_{+}(x)-{(f+g)}_{\text{−}}(x)\stackrel{\text{!}}{\text{=}}(f_{\text{+}}(x)-f_{\text{−}}(x))+(g_{\text{+}}(x)-g_{\text{−}}(x))$$

  or in other words,

  $$\max <!--\; FUNCTION\; APPLICATION\; -->\{f(x)+g(x),0\}-\max <!--\; FUNCTION\; APPLICATION\; -->\{-f(x)-g(x),0\}\stackrel{\text{!}}{\text{=}}\max <!--\; FUNCTION\; APPLICATION\; -->\{f(x),0\}-\max <!--\; FUNCTION\; APPLICATION\; -->\{-f(x),0\}+\max <!--\; FUNCTION\; APPLICATION\; -->\{g(x),0\}-\max <!--\; FUNCTION\; APPLICATION\; -->\{-g(x),0\}$$

  is indeed valid. We look at the possible cases for an arbitrary $x\in {ℝ}^d$, in all cases we try to transform $\max <!--\; FUNCTION\; APPLICATION\; -->\{f(x)+g(x),0\}-\max <!--\; FUNCTION\; APPLICATION\; -->\{-f(x)-g(x),0\}$

  1)

  $f(x)+g(x)\ge 0$. From this we already have $\max <!--\; FUNCTION\; APPLICATION\; -->\{f(x)+g(x),0\}-\max <!--\; FUNCTION\; APPLICATION\; -->\{-f(x)-g(x),0\}=$ $\max <!--\; FUNCTION\; APPLICATION\; -->\{f(x)+g(x),0\}$. This induces further cases

  1.1)

  $f(x)\ge 0$ and $g(x)\le 0$. Then $$\begin{array}{llll}\hfill \max <!--\; FUNCTION\; APPLICATION\; -->\{f(x)+g(x),0\}& =\max <!--\; FUNCTION\; APPLICATION\; -->\{f(x),0\}-\max <!--\; FUNCTION\; APPLICATION\; -->\{-g(x),0\}& \hfill & \\ \hfill & =\max <!--\; FUNCTION\; APPLICATION\; -->\{f(x),0\}-\max <!--\; FUNCTION\; APPLICATION\; -->\{-f(x),0\}+\max <!--\; FUNCTION\; APPLICATION\; -->\{g(x),0\}-\max <!--\; FUNCTION\; APPLICATION\; -->\{-g(x),0\}& \hfill & \end{array}$$

  as desired.

  1.2)

  $f(x)\ge 0$ and $g(x)>0$. Then $$\begin{array}{llll}\hfill \max <!--\; FUNCTION\; APPLICATION\; -->\{f(x)+g(x),0\}& =\max <!--\; FUNCTION\; APPLICATION\; -->\{f(x),0\}+\max <!--\; FUNCTION\; APPLICATION\; -->\{g(x),0\}& \hfill & \\ \hfill & =\max <!--\; FUNCTION\; APPLICATION\; -->\{f(x),0\}-\max <!--\; FUNCTION\; APPLICATION\; -->\{-f(x),0\}+\max <!--\; FUNCTION\; APPLICATION\; -->\{g(x),0\}-\max <!--\; FUNCTION\; APPLICATION\; -->\{-g(x),0\}& \hfill & \end{array}$$

  as desired.

  1.3)

  $f(x)<0$ and $g(x)\ge 0$ follows similarly to 1.1

  1.4)

  $f(x)<0$ and $g(x)<0$ is impossible.

  2)

  The case for $f(x)-g(x)<0$ follows symmetrically.

  We now demonstrate the scalar multiplication property. We consider several cases:

  • $c\ge 0$
    Notice that for any real number $y$ we then have $\max <!--\; FUNCTION\; APPLICATION\; -->\{\mathit{cy},0\}=c\max <!--\; FUNCTION\; APPLICATION\; -->\{y,0\}$, and thus

    $$\begin{array}{llll}\hfill \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\mathit{cf}& =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{(\mathit{cf})}_{\text{+}}-\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{(\mathit{cf})}_{\text{−}}=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}c{(f)}_{\text{+}}-\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}c{(f)}_{\text{−}}& \hfill & \\ \hfill & =c\cdot \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}_{\text{+}}-c\cdot \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}_{\text{−}}=c\cdot \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f& \hfill & \end{array}$$

  • $c=-1$
    We then have ${(-f)}_{\text{+}}=\max <!--\; FUNCTION\; APPLICATION\; -->\{-f,0\}=f_{\text{−}}$ and similarly ${(-f)}_{\text{−}}=f_{\text{+}}$; thus,

    $$\begin{array}{llll}\hfill \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}-f& =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{(-f)}_{\text{+}}-\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{(-f)}_{\text{−}}=\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}_{\text{−}}-\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}_{\text{+}}& \hfill & \\ \hfill & =-1\cdot \left(\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}_{\text{+}}-\cdot \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}{f}_{\text{−}}\right)=-\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f& \hfill & \end{array}$$
  • $c<0$
    Follows by combining the first and the second case.
• (Equivalence)
  Notice that $f\stackrel{a.e.}{\text{=}}g$ trivially implies that $f_{\text{+}}\stackrel{a.e.}{\text{=}}{g}_{\text{+}}$ and $f_{\text{−}}\stackrel{a.e.}{\text{=}}{g}_{\text{−}}$ (for instance by $\{x\in {ℝ}^d:f_{\text{+}}(x)\ne g_{\text{+}}(x)\}\subseteq \{x\in {ℝ}^d:f(x)\ne g(x)\}$). The equality of the intervals then follows directly from [?](iv) by applying it to both positive and negative part of the real valued simple integral.
• (Compatibility with the Lebesgue measure)
  Since the indicator function is non-negative, this follows trivially from [?](vi)

Complex-valued simple integral.

• (Linearity)
  Since the complex simple integral can be written in terms of two real valued functions

  $$\begin{array}{llll}\hfill \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f+g& =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Re (f+g)+i\cdot \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Im (f+g)& \hfill & \\ \hfill & =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}[\Re (f)+\Re (g)]+i\cdot \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}[\Im (f)+\Im (g)]& \hfill & \end{array}$$

  Notice that both $\Re (f)+\Re (g)$ and $\Im (f)+\Im (g)$ are real valued functions; applying the previous part of the proof we get the desired result.
  The scalar multiplication property is demonstrated as follows. We can rewrite $\mathit{cf}$ as

  $$\mathit{cf}(x)=[\Re (c)+\mathit{i\Im }(c)]\cdot [\Re (f(x))+\mathit{i\Im }(f(x))]=\Re (c)\cdot \Re (f(x))-\Im (c)\Im (f)+\mathit{i\Im }(c)\Re (f)+\mathit{i\Re }(c)\Im (f)$$

  Using this identity we obtain

  $$\begin{array}{llll}\hfill \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\mathit{cf}& =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Re (\mathit{cf})+i\cdot \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Im (\mathit{cf})& \hfill & \\ \hfill & =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}[\Re (c)\Re (f)-\Im (c)\Im (f)]+i\cdot \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}[\Im (c)\Re (f)+\Re (c)\Im (f)]& \hfill & \\ \hfill & =\Re (c)\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Re (f)-\Im (c)\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Im (f)+\mathit{i\Im }(c)\cdot \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Re (f)+\Re (c)\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Im (f)& \hfill & \\ \hfill & =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Re (f)\cdot \left[\Re (c)+\mathit{i\Im }(c)\right]+i\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Im (f)\left[\Re (c)-1∕\mathit{i\Im }(c)\right]& \hfill & \\ \hfill & =\left[\Re (c)+\mathit{i\Im }(c)\right]\cdot \left(\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Re (f)+i\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Im (f)\right)& \hfill & \\ \hfill & =c\cdot \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f& \hfill & \end{array}$$

  The conjuctive property follows as follows

  $$\begin{array}{llll}\hfill \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\overline{f}& =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Re (\overline{f})+i\cdot \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Im (\overline{f})& \hfill & \\ \hfill & =\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Im (f)+i\cdot \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Re (f)& \hfill & \\ \hfill & =\overline{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Re (f)+i\cdot \mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}\Im (f)}& \hfill & \\ \hfill & =\overline{\mathrm{Simp}{\int \; <!--\; nolimits\; -->}_{{ℝ}^d}f}& \hfill & \end{array}$$
• (Equivalence)
  Equivalence follows by applying the analogous property of the real valued functions separately to the real and imaginary part of the complex simple integral.

Let $f={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=1}^m{c}_i{\text{⊮}}_{E_i}(x)$ be an absolutely integrable simple function and let $\phi :{\text{Simp}}^{\text{abs}}({ℝ}^d)\to ℂ$ be a mapping such that the above properties are given.

First, let’s restrict ourselves to real-valued functions to see that $\phi$ equates to ${\text{Simp}}^{ℝ}{\int }_{{ℝ}^d}f(x)\mathit{dx}$. For this, define $M=\{1,\dots ,m\}$ and $M_{\text{+}}=\{i\in M:c_i\ge 0\}$ and $M_{\text{−}}=\{i\in M:c_i<0\}$. This is a partition of $M$.

Hence, $\phi$ agrees with the integral. Now assume $f$ is complex-valued, then

Thus, $\phi$ is equal to the complex-valued simple integral, and we are done.