Exercise 1.3.3 (Examples of unsigned Lebesgue measurable functions)

Question

Show that\begin{enumerate}[label=(oman*)]
\item every unsigned continuous function $f:\mathbb{R}^d	o[0,\infty]$ is unsigned Lebesgue measurable.\item every unsigned simple function $f:\mathbb{R}^d	o[0,\infty]$ is unsigned Lebesgue measurable.\item supremum, infimum, limit superior, limit inferior of unsigned Lebesgue measurable functions is unsigned Lebesgue measurable.\item an unsigned function that is equal almost everywhere to an unsigned Lebesgue measurable function, is itself unsigned Lebesgue measurable.\item if a sequence $(f_n)_{n\in\mathbb{N}}$ of unsigned Lebesgue measurable functions converges pointwise almost everywhere to an unsigned limit $f$ then $f$ is also unsigned Lebesgue measurable.\item if $f:\mathbb{R}^d	o[0,\infty]$ unsigned Lebesgue measurable function and $\phi:[0,+\infty] 	o [0,+\infty]$ is continuous, then $\phi \circ f$ is unsigned Lebesgue measurable.\item if $f,g$ are unsigned Lebesgue measurable functions, then $f+g$ and $fg$ are also unsigned Lebesgue measurable functions.\end{enumerate}

Elu Thingol · Accepted Answer

\begin{enumerate}[label=(oman*)]
\item Let $f$ be an unsigned continuous function. By Theorem 2.1.5 (see Prof. Tao's \href{../analysis_II}{Analyis II}) the inverse image of every open set is open, and thus measurable. By Lemma 1.3.9 (x) $f$ is Lebesgue measurable.\item Let $f$ be an unsigned simple function. Then $f$ is trivially a pointwise limit of a constant sequence consisting of itself, and is thus Lebesgue measurable.\item We use the criteria (v) from Lemma 1.3.9. Pick an arbitrary $\lambda \geq 0$. In other words, we want to demonstrate that
$$\left\{x \in \mathbb{R}^d: \sup_{n\in\mathbb{N}}f_n(x) \geq \lambda ight\}$$
is Lebesgue measurable. But this follows directly from the identity
$$\left\{ x \in \mathbb{R}^d: \sup_{n\in\mathbb{N}}f_n(x) \geq \lambda ight\} = \bigcup_{n\in\mathbb{N}} \left\{ x \in \mathbb{R}^d: f_n(x) \geq \lambda ight\}$$
by the Lebesgue measurability of $f_n$'s and the invariance of Lebesgue measurability under the countable unions. Thus, we prove the above identity in what follows.
	\begin{itemize}
	\item[$\subseteq$)] Let $y \in \left\{ x \in \mathbb{R}^d: \sup_{n\in\mathbb{N}}f_n(x) \geq \lambda ight\}$ arbitrary. Notice that $\forall n \in \mathbb{N}$ the value $f_n(y) \leq \sup_{n\in\mathbb{N}}f_n(x)$.
	\item[$\supseteq$)] Instead of showing $A \subseteq B$ we demonstrate the equivalent $\mathbb{R}^d/A \supseteq \mathbb{R}^d/B$. Thus, let $y \in \left\{ x \in \mathbb{R}^d: \sup_{n\in\mathbb{N}}f_n(x) < \lambda ight\}$. We then conclude that there is no $n \in \mathbb{N}$ such that $f_n(x) \geq \lambda$, and hence $y
otin \bigcup_{n\in\mathbb{N}} \left\{ x \in \mathbb{R}^d: f_n(x) \geq \lambda ight\}$.
	\end{itemize}
	
\item Let $f:\mathbb{R}^d	o[0,\infty]$ be unsigned Lebesgue measurable function with a sequence $(f_n)_{n\in\mathbb{N}}$ of unsigned measurable functions. Let $g:\mathbb{R}^d	o[0,\infty]$ be an unsigned function such that almost everywhere $f=g$. Let $N\subset \mathbb{R}^d$ be the null set on which $f$ and $g$ disagree. Since $g$ is simple, it admits some simple values $c_1,\ldots,c_k$ on some $E_1,\ldots,E_k\subseteq \mathbb{R}^d$. Consider a sequence $N\cap E_1,\ldots, N\cap E_k$ and $c_{k+1}=0$ for $N/E_{k+} := N / (E_1\cup \ldots \cup E_k)$, and define$$\forall n \in \mathbb{N}: \quad g_n(x) = \left\{\begin{array}{ll} f_n(x) &	ext{ if $x
otin N$} \ c_i \cdot &	ext{ if $x \in N\cap E_i$}\end{array}ight.$$      Then we obviously have $f_n(x) 	o f(x)$ on both $\mathbb{R}^d/N$ and $N$.

\item Let $(f_n)_{n\in\mathbb{N}} {	o} f$ be a sequence of unsigned Lebesgue measurable functions (we got rid of almost everywhere notion by the equivalence of (ii) and (iii) in Lemma 1.3.9).
Since limit of a function is in particular its limit inferior (or alternatively limit superior):
$$\forall x \in \mathbb{R}^d: \quad f(x)=\lim_{n	o\infty} f_n(x) = \limsup_{n	o\infty}f_n(x) = \inf_{N\in\mathbb{N}} \sup_{n \geq N} f_n(x)$$
By part (iii) for any $N\in\mathbb{N}$, the pointwise supremum $\sup_{n\geq N} f_n$ is unsigned Lebesgue measurable. By part (iii) again, the same applies to infimum, and the limit must thus be unsigned Lebesgue measurable.\item We use the criteria (x) from Lemma 1.3.9. Let $U \subseteq [0,+\infty]$ be an open set. Then using some set theory we see that
$$(\phi\circ f)^{-1}(U) = f^{-1}(\phi^{-1}(U))$$
Since $\phi$ is continuous, $\phi^{-1}(U)\subseteq \mathbb{R}$ must be open as well. By Lemma 1.3.9 the set $f^{-1}(\phi^{-1}(U))$ must be measurable, and so the function $\phi\circ f$ must be unsigned Lebesgue measurable.

\item Let $(f_n)_{n\in\mathbb{N}}$ and $(g_n)_{n\in\mathbb{N}}$ be sequences of unsigned Lebesgue measurable sets converging pointwise to $f$ and $g$ respectively. Then, $(f_n+g_n)_{n\in\mathbb{N}}$ and $(f_ng_n)_{n\in\mathbb{N}}$ must converge pointwise to $f+g$ and $fg$ by limit laws, so both must be unsigned Lebesgue measurable.
\end{enumerate}

Exercise 1.3.3 (Examples of unsigned Lebesgue measurable functions)

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Exercise 1.3.3 (Examples of unsigned Lebesgue measurable functions)

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