Exercise 1.3.8 (Examples of complex Lebesgue measurable functions)

Question

Show that
\begin{enumerate}[label=(oman*)]
\item Every complex continuous function $f:\mathbb{R}^d	o\mathbb{C}$ is Lebesgue measurable\item A function $f:\mathbb{R}^d	o\mathbb{C}$ is simple if and only if it is Lebesgue measurable and takes on at most finitely many values.\item A complex-valued function that is equal almost everywhere to an measurable function, is itself measurable.\item If a sequence $(f_n)_{n\in\mathbb{N}}$ of (complex) Lebesgue measurable functions converges pointwise almost everywhere to an complex-valued limit $f$, then $f$ is also measurable.\item If $f:\mathbb{R}^d	o\mathbb{C}$ is measurable and $\phi:\mathbb{C}	o\mathbb{C}$ is continuous, then $\phi \circ f$ is measurable.\item If $f,g$ are Lebesgue measurable functions, then $f+g$ and $fg$ are Lebesgue measurable as well.\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(oman*)]
\item By criteria (iv) from \href{./exercise_1-3-7}{Exercise 1.3.7} $f$ is measurable, since the pre-image of any open set under the continuous function is open, and thus Lebesgue measurable.
\item Suppose $f:\mathbb{R}^d	o\mathbb{C}$ is simple with the collection of values $c_1,\ldots,c_n \in \mathbb{C}$ associated with the sets $E_1,\ldots,E_n \subseteq \mathbb{R}^d$. By definition it takes on $n \in \mathbb{N}$ values. Furthermore, it is the limit of the constant sequence $(f)_{n\in\mathbb{N}}$ consisting of itself, so it must be Lebesgue measurable. Now suppose conversely that $f:\mathbb{R}^d	o\mathbb{C}$ is Lebesgue measurable and takes on at most finitely many values, $c_1,\ldots,c_n \in \mathbb{C}$. Then for any $1\leq i \leq n$ we have $f^{-1}(c_i)$ is necessarily closed, since $\{c_i\}$ is a closed set in $\mathbb{C}$, and the inverse image of a closed set is measurable under the Lebesgue measurable function $f$. Thus, $E_1 = f^{-1}(c_1),\ldots, E_n=f^{-1}(c_n)$ are Lebesgue measurable, and $f$ is simple.\item Suppose that $g:\mathbb{R}^d	o\mathbb{C}$ is a function such that $g = f$ almost everywhere for a Lebesgue measurable function $f:\mathbb{R}^d	o\mathbb{C}$. Let $N_1\subset \mathbb{R}^d$ be a null set on which $f|_N 
eq g|_N$. By definition, we can find a sequence of simple functions $(f_n)_{n\in\mathbb{N}}$ such that this sequence converges to $f$ everywhere except for a null set $N_2$. The set $N_1 \cup N_2$ is still a null set, and we have
$$\forall x \in \mathbb{R}^d: \quad (f_n(x))_{n\in\mathbb{N}} 	o f(x) = g(x)$$
In other words, we have found a sequence of simple functions which converge pointwise to $g$ almost everywhere, as desired.\item Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of complex Lebesgue measurable functions converging to $f$.
By \href{./exercise_1-3-7}{Exercise 1.3.7} (iii) this implies that the sequences $(\Re f_n)^+_{n\in\mathbb{N}}$, $(\Re f_n)^-_{n\in\mathbb{N}}$, $(\Im f_n)^+_{n\in\mathbb{N}}$ and $(\Im f_n)^-_{n\in\mathbb{N}}$ are sequences of unsigned Lebesgue measurable functions. We have proven this assertion for unsigned Lebesgue measurable functions in \href{./exercise_1-3-3}{Exercise 1.3.3} (iv); thus all of the above sequences converge to an unsigned Lebesgue measurable function. But then, the functions
\begin{align*}
\Re f &= \lim^{p}_{n	o\infty} \Re f_n = \lim^{p}_{n	o\infty} (\Re f_n)^+ - \lim^{p}_{n	o\infty} (\Re f_n)^-\
\Im f &= \lim^{p}_{n	o\infty} \Im f_n = \lim^{p}_{n	o\infty} (\Im f_n)^+ - \lim^{p}_{n	o\infty} (\Im f_n)^-
\end{align*}
must be signed Lebesgue measurable, as combinations of unsigned Lebesgue measurable functions. Finally, $f = \Re f + i \cdot \Im f$ must be complex Lebesgue measurable.\item Let $U \subseteq \mathbb{C}$ be an open set. By definition, the inverse image $\phi^{-1}(U) \subseteq \mathbb{C}$ must be open as well. Finally, $f^{-1}(\phi^{-1}(U)) = (\phi \circ f)^{-1}(U)$ is Lebesgue measurable; by criteria (iv) from \href{./exercise_1-3-7}{Exercise 1.3.7}, $\phi \circ f$ must be Lebesgue measurable.\item If $(f_n)_{n\in\mathbb{N}}$ is a sequence of complex simple functions converging pointwise almost everywhere to $f$, and $(g_n)_{n\in\mathbb{N}}$ is a sequence of complex simple functions converging pointwise almost everywhere to $g$, then by limit laws for complex numbers the sequence of complex simple functions $(f_n+g_n)_{n\in\mathbb{N}}$ must converge to $f+g$ and we are done. An identical argument gives us the Lebesgue measurability of $fg$.
\end{enumerate}

Exercise 1.3.8 (Examples of complex Lebesgue measurable functions)

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Exercise 1.3.8 (Examples of complex Lebesgue measurable functions)

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