Exercise 1.3.7 (Equivalent formulations of Lebesgue measurability of functions)

In the following we will often treat $ℂ$ as ${ℝ}^2$.

• $(i)⟺(\mathit{ii})$
  By definition.

• $(\mathit{ii})⟺(\mathit{iii})$
  Let ${(f_n)}_{n\in \mathbb{N}}$ be a sequence of complex functions which converge to $f$ poinwise almost everywhere. We only consider $r^{\text{+}}:=\max <!--\; FUNCTION\; APPLICATION\; -->\{\Re (f),0\}$; the others $\min <!--\; FUNCTION\; APPLICATION\; -->\{\Re (f),0\},\max <!--\; FUNCTION\; APPLICATION\; -->\{\Im (f),0\},\min <!--\; FUNCTION\; APPLICATION\; -->\{\Im (f),0\}$ follow in analogous manner. First, it is easy to verify by cases that we have the pointwise convergence of the positive parts:

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->x\in {ℝ}^d:\max <!--\; FUNCTION\; APPLICATION\; -->\{\Re (f_n(x)),0\}\to \max <!--\; FUNCTION\; APPLICATION\; -->\{\Re (f(x)),0\}$$

  Furthermore, each $\max <!--\; FUNCTION\; APPLICATION\; -->\{\Re (f_n),0\}$ is simple, since $\Re (f_n)$ is simple with finite number of values $\{c_1,\dots ,c_n\}$, and $\max <!--\; FUNCTION\; APPLICATION\; -->\{\Re (f_n),0\}$ simply takes on the positive ones. Thus, we have constructed a sequence ${(\max <!--\; FUNCTION\; APPLICATION\; -->\{\Re (f_n),0\})}_{n\in \mathbb{N}}$ of simple functions which converge pointwise to $\max <!--\; FUNCTION\; APPLICATION\; -->\{\Re (f),0\}$.
  For the converse statement, find sequences of unsigned simple functions ${(r_n^{\text{+}})}_{n\in \mathbb{N}},{(r_n^{\text{−}})}_{n\in \mathbb{N}},{(j_n^{\text{+}})}_{n\in \mathbb{N}},{(j_n^{\text{−}})}_{n\in \mathbb{N}}$ which converge to the positive and negative parts of $\Re (f)$ and $\Im (f)$ respectively. Set

  $$\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}:f_n:=r_n^{\text{+}}+r_n^{\text{−}}+j_n^{\text{+}}+j_n^{\text{−}}$$

  We then have by limit laws

  $$\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}{f}_n(x)=\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}(r_n^{\text{+}}(x)+r_n^{\text{−}}(x))+\underset{n\to \infty }{\lim <!--\; FUNCTION\; APPLICATION\; -->}(j_n^{\text{+}}(x)+j_n^{\text{−}}(x))=\Re (f)(x)+\Im (f)(x)=f(x)$$

• $(\mathit{iv})⟺(v)$
  Let $K\subseteq ℂ$ be a closed set. Then $ℂ∕K$ is open, and

  $$f^{-1}(ℂ∕K)=\left\{x\in {ℝ}^d:f(x)\notin K\right\}={ℝ}^d∕\left\{x\in {ℝ}^d:f(x)\in K\right\}={ℝ}^d∕f^{-1}(K)$$

  is Lebesgue measurable. But then $f^{-1}(K)$ must also be Lebesgue measurable by the complementarity property from Lemma 1.2.13 (v). A symmetric argument demonstrates the converse statement.

• $(\mathit{iii})⟺(\mathit{iv})$ for signed Lebesgue measurable functions

  • Let $f:{ℝ}^d\to ℝ$ be a signed Lebesgue measurable function, and assume that for any open set $U$, $f^{-1}(U)$ is measurable. Now consider the positive part $(f^{\text{+}})$ and the negative part $(f^{\text{−}})$ of $f$. Our goal is to show that for any positive open set $U\subseteq [0,+\infty )$ (notice that $0\notin U$) the inverse images ${(f^{\text{+}})}^{-1}(U)$ and ${(f^{\text{−}})}^{-1}(U)$ are open. But this is trivial due to:

    $$\begin{array}{llll}\hfill {(f^{\text{+}})}^{-1}(U)& =\left\{x\in {ℝ}^d:(f^{\text{+}})(x)\in U\right\}& \hfill & \\ \hfill & =\left\{x\in {ℝ}^d:f(x)\in (0,+\infty )\text{ and }f(x)\in U\right\}& \hfill & \\ \hfill & =\left\{x\in {ℝ}^d:f(x)\in (0,+\infty )\right\}\cap \left\{x\in {ℝ}^d:f(x)\in U\right\}& \hfill & \\ \hfill & =\left\{x\in {ℝ}^d:f(x)\in U\right\}=f^{-1}(U)& \hfill & \end{array}$$

    where $f^{-1}(U)$ is Lebesgue measurable by assumption. A similar argument gives us the unsigned Lebesgue measurablity of ${(f^{\text{−}})}^{-1}$:

    $$\begin{array}{llll}\hfill {(f^{\text{−}})}^{-1}(U)& =\left\{x\in {ℝ}^d:(f^{\text{−}})(x)\in U\right\}& \hfill & \\ \hfill & =\left\{x\in {ℝ}^d:-f(x)\in (0,+\infty )\text{ and }-f(x)\in U\right\}& \hfill & \\ \hfill & =\left\{x\in {ℝ}^d:f(x)\in (-\infty ,0)\right\}\cap \left\{x\in {ℝ}^d:f(x)\in -U\right\}& \hfill & \\ \hfill & =\left\{x\in {ℝ}^d:f(x)\in -U\right\}=f^{-1}(-U)& \hfill & \end{array}$$

    Due to the fact that the openness of $U$ implies the openness of $-U$, the Lebesgue measurability follows.

  • Now assume that for our $f$, both positive parts $f^{\text{+}}$ and negative parts $f^{\text{−}}$ are unsigned Lebesgue measurable. First of all, this implies that the inverse image of any half-open interval $J=(-\infty ,\lambda )$ is measurable. This is easy to verify, due to the fact that in the sum $f(x)=f^{\text{+}}(x)-f^{\text{−}}(x)$ at least one of the terms will always be zero:

    $$f^{-1}(J)=(f^{\text{+}}-f^{\text{−}})(J)=\left\{x\in {ℝ}^d:f^{\text{+}}(x)-f^{\text{−}}(x)<\lambda \right\}=\left\{x\in {ℝ}^d:f^{\text{+}}(x)<\lambda \right\}\cup \left\{x\in {ℝ}^d:-f^{\text{−}}(x)<\lambda \right\}$$

    But both of the above sets are the inverse images of the open set $J=(-\infty ,\lambda )$ under the unsigned Lebesgue measurable $f^{\text{+}}$ and $f^{\text{−}}$; by Lemma 1.3.9 (vii) $f^{-1}(I)$ must be Lebesgue measurable.
    This can be easily extended to an arbitrary interval by observing that for $a,b\in ℝ$ the interval $I=(a,b)$ can be represented as $(-\infty ,b)∕(-\infty ,a)$; and thus,

    $$f^{-1}\left(I\right)=f^{-1}\left((-\infty ,b)∕(-\infty ,a)\right)=f^{-1}\left((-\infty ,b)\right)∕f^{-1}\left((-\infty ,a)\right)$$

    which by Lemma 1.2.13 (v) is also Lebesgue measurable.
    Finally, any open subset $U\subseteq ℝ$ can be represented as countable union of disjoint open intervals ${(I_n)}_{n\in \mathbb{N}}$; with this we finally conclude

    $$f^{-1}(U)=f^{-1}\left(\bigcup _{n\in \mathbb{N}}{I}_n\right)=\bigcup _{n\in \mathbb{N}}{f}^{-1}(I_n)$$

    and $f^{-1}(U)$ is Lebesgue measurable by Lemma 1.2.13 (vi).

• $(\mathit{iii})⟺(\mathit{iv})$ for complex Lebesgue measurable functions.

  • Suppose that for any open set we have the Lebesgue measurability of its inverse under $f$. Let $V$ be an open set in $[0,+\infty )$; we would like to demonstrate that ${(\mathit{\Re f})}^{\text{+}},{(\mathit{\Re f})}^{\text{−}},{(\mathit{\Im f})}^{\text{+}}$ and ${(\mathit{\Im f})}^{\text{−}}$ return a Lebesgue measurable set when taken inverse under $V$. Using the same logic as in the proof for real-valued $f$ we get

    $$\begin{array}{llll}\hfill {({(\mathit{\Re f})}^{\text{+}})}^{-1}(V)& ={(\mathit{\Re f})}^{-1}(V)=f^{-1}({\Re }^{-1}(V))& \hfill & \\ \hfill {({(\mathit{\Re f})}^{\text{−}})}^{-1}(V)& ={(\mathit{\Re f})}^{-1}(-V)=f^{-1}({\Re }^{-1}(-V))& \hfill & \end{array}$$

    Since the projection $\Re$ is a continuous function, ${\Re }^{-1}(V)$ and ${\Re }^{-1}(-V)$ must also be open. The inverse of an open set under $f$ is Lebesgue measurable by assumption, and we are done. The same argument gives us the positive and the negative magnitude of the imaginary part.

  • Now suppose that ${(\mathit{\Re f})}^{\text{+}},{(\mathit{\Re f})}^{\text{−}},{(\mathit{\Im f})}^{\text{+}}$ and ${(\mathit{\Im f})}^{\text{−}}$ are unsigned Lebesgue measurable. Then $\mathit{\Re f}={(\mathit{\Re f})}^{\text{+}}-{(\mathit{\Re f})}^{\text{−}}$ and $\mathit{\Im f}={(\mathit{\Im f})}^{\text{+}}-{(\mathit{\Im f})}^{\text{−}}$ must be signed Lebesgue measurable. Notice that we can represent $f:{ℝ}^d\to ℂ$ with some mapping $x\mapsto (a,b)$ as a coordinatewise function

    $$x\mapsto (\mathit{\Re f}(x),\mathit{\Im f}(x))$$

    We mentally verify that we can then write for any $(a,b)\in ℂ$:

    $$f^{-1}(\{(a,b)\})=\left\{x\in {ℝ}^d:x\in {(\mathit{\Re f})}^{-1}(a)\cap {(\mathit{\Im f})}^{-1}(b)\right\}$$

    Let $U\subseteq ℂ$ be an arbitrary open set. Then

    $$f^{-1}(U)=\bigcup _{(a,b)\in U}\left\{x\in {ℝ}^d:x\in {(\mathit{\Re f})}^{-1}(a)\cap {(\mathit{\Im f})}^{-1}(b)\right\}={(\mathit{\Re f})}^{-1}({\pi }_1(U))\cap {(\mathit{\Im f})}^{-1}({\pi }_2(U))$$

    Where ${\pi }_1$ and ${\pi }_2$ denote the projections of $U$ w.r.t. its first and second coordinate. From topology we know that projection is an open map; thus ${\pi }_1(U)$ and ${\pi }_2(U)$ are both open. We have proven this theorem for real-valued Lebesgue measurable functions; in other words, $\mathit{\Re f}$, and $\mathit{\Im f}$ will return Lebesgue measurable sets when taken inverse under an open set. We hence conclude that $f^{-1}(U)$ is a Lebesgue measurable set, since it is an intersection of two Lebesgue measurable sets.