Ex4-6

Question: Let f and g be measurable functions on E.
(a) If f and g are finite a.e. in E, show that f + g is measurable no matter how we define it at the points when it has the form +\infty + (-\infty) or -\infty + \left ( +\infty \right ) .
(b)Show that fg is measurable without restriction on the finiteness of f and g. Show that f + g is measurable if it is defined to have the same value at every point where it has the form +\infty + (-\infty) or -\infty + \left ( +\infty \right ) .
(Note that a function h defined on E is measurable if and only if both \left \{ h=+\infty \right \} and \left \{ h=-\infty \right \} are measurable and the restriction of h to the subset of E where h is finite is measurable.) 
Solution:
(a) Recall the following theorem: 
Let f:E\to \mathbb{R} \cup \left \{ \infty ,-\infty \right \} and g:E\to\mathbb{R} \cup \left \{ \infty ,-\infty \right \} be two functions satisfying that f is measurable and g(x) = f(x) for almost every x\in E . Then, g(x) is measurable.
We define a new function h(x) such that h(x)=\left\{\begin{array}{ll}
g(x) & \text { if } g(x) \in \mathbb{R} \\
0 & \text { if } g(x) \in\{\pm \infty\}
\end{array}\right. .
Since g is measurable and g(x)=h(x) almost everywhere, h(x) is measurable. Now, it remains to show f+h is measurable.
Now, we have f(x)+h(x)=\left\{\begin{array}{ll}
f(x)+g(x) & \text { if } g(x) \in \mathbb{R} \\
f(x) & \text { if } g(x) \in\{\pm \infty\}
\end{array}\right. .
Consider the set \left \{ f+h >a \right \}=(\left \{ f>a \right \}\cap\left \{ g= \pm \infty\right \})\cup (\left \{ f+g>a \right \}\cap\left \{ g\ne \pm \infty\right \}) . 
\left \{ f>a \right \} and \left \{ g= \pm \infty\right \} is measurable obviously. And \left \{ g \ne \pm \infty\right \} is measurable since it is complement of A. Next, we will claim that \left \{ f+g>a \right \} is measurable as well.
Let us write Q as Q=\left \{ r_{1},r_{2},r_{3}... \right \} , since Q is countable. Then we could write \left \{ f+g>a \right \} as
 \begin{aligned}
\left \{ f+g>a \right \}&= \bigcup_{k=1}^{\infty} \left \{ f>r_{k}>-g+a \right \} \\
&=\bigcup_{k=1}^{\infty} (\left \{ f>r_{k} \right \})\cap(\left \{ -g+a<r_{k} \right \})
 \end{aligned}
 
Since f and g are both measurable, both 
 \left \{ f>r_{k} \right \}
 and 
 \left \{ -g+a<r_{k} \right \}
 are measurable, which implies that 
 \left \{ f+g>a \right \}
 is measurable.
Combining all these, we can conclude that \left \{ f+h >a \right \} is measurable, which means the function f+h:E \to \mathbb{R}^{n} \cup \left \{ \infty,-\infty \right \} is measurable.
Since f+h is measurable and (f+g)(x)=(f+h)(x) almost everywhere, we have f+g is measurable.
----(b) We first prove two lemmas.
Lemma 1: Let E_1, E_2 \sube \R^n be measurable sets and f: E_1 \cup E_2 \to \R be a function defined on E_1 \cup E_2. Then f is measurable if and only if its restrictions to E_1 (f|_{E_1} = f: E_1 \to \mathbb{R}) and E_2 (f|_{E_2} = f: E_2 \to \R) are measurable. 
Proof: Suppose that f is measurable. Then for any a \in \R, the set\{\mathbf{x} \in E_1 \cup E_2 \ | \ f(\mathbf{x}) > a\}is measurable. Fix a \in \R. Then\{\mathbf{x} \in E_1 \mid f_{E_1}(\mathbf{x}) > a\} = \{\mathbf{x} \in E_1 \cup E_2 \mid f(\mathbf{x}) > a\} \cap E_1which is measurable. Likewise, we show that \{\mathbf{x} \in E_2 \mid f_{E_2}(\mathbf{x}) > a\} is measurable. Therefore, we conclude that f|_{E_1} and f|_{E_2} are measurable. 
Conversely, suppose that f|_{E_1} and f|_{E_2} are measurable. Then by the formula\begin{align}
 \{\mathbf{x} \in E_1 &\cup E_2 \mid f(\mathbf{x}) > a\} \\
 &= \{\mathbf{x} \in E_1 \mid f_{E_1}(\mathbf{x}) > a\} \cup \{\mathbf{x} \in E_2 \mid f_{E_2}(\mathbf{x}) > a\},
\end{align}we see immediately that f is measurable.
Lemma 2: Let E \sube \R^n be measurable and f: E \to\overline{\R} is a function defined on E, where \overline{\R} = \R \cup \{-\infty, \infty\} is the extended real line. If E_1 = \{f(\mathbf{x}) = \infty\} and E_2 = \{f(\mathbf{x}) = - \infty\}, then f is measurable if and only if E_1 and E_2 are measurable and the restriction f|_{E \setminus (E_1 \cup E_2)} is measurable.
Proof: Suppose that f is measurable. Then E_1 and E_2 are measurable by Corollary 4.2 (from the textbook). It follows that E \setminus (E_1 \cup E_2 ) is measurable, and so is the restriction f|_{E \setminus (E_1 \cup E_2)} by Lemma 1. 
Conversely, suppose that E_1 and E_2 are measurable and the restriction f|_{E \setminus (E_1 \cup E_2)} is measurable. Then for any a \in \R, we have \{\mathbf{x} \in E \mid f(\mathbf{x}) > a\} = E_1 \cup \{\mathbf{x} \in E \mid f|_{E \setminus (E_1 \cup E_2)}(\mathbf{x}) > a\}which is measurable, and so is f. 
We first show that fg is measurable without the restriction on finiteness of f and g. Let E_1 \{fg = \infty\} \qquad \text{and} \qquad E_2 = \{fg = - \infty\}. 
Then \begin{align}
E_1 &= \{\mathbf{x} \in E \mid f(\mathbf{x}) = \infty, g(\mathbf{x}) > 0\} \cup \{\mathbf{x} \in E \mid f(\mathbf{x}) = - \infty, g(\mathbf{x}) < 0\} \\ 
&\quad \cup \{\mathbf{x} \in E \mid f(\mathbf{x}) > 0, g(\mathbf{x}) = \infty\} \cup \{\mathbf{x} \in E \mid f(\mathbf{x}) < 0, g(\mathbf{x}) = - \infty\},
\end{align}which is measurable. Likewise, we show that E_2 is measurable. By Lemma 2, it suffices to show that h = fg|_{E \setminus (E_1 \cup E_2)} is measurable. Given a \in \R. If a \ge 0, then\begin{align}
&\{\mathbf{x} \in E \setminus (E_1 \cup E_2) \mid h(\mathbf{x}) > a\} \\
&= \left\{\mathbf{x} \in E \setminus (E_1 \cup E_2) \ \big| \ \left(f|_{E \setminus \{f \pm \infty\}}(\mathbf{x}) \right) \cdot \left(g|_{E \setminus \{g \pm \infty\}}(\mathbf{x}) \right) > a \right\},
\end{align}which is measurable by Theorem 4.10 (in the textbook). 
If a < 0, then \begin{align}
&\{\mathbf{x} \in E \setminus (E_1 \cup E_2) \mid h(\mathbf{x}) > a\} \\
&= \{\mathbf{x} \in E \setminus (E_1 \cup E_2) \mid f(\mathbf{x}) = 0\} \cup \{\mathbf{x} \in E \setminus (E_1 \cup E_2) \mid g(\mathbf{x}) = 0\} \\
&\quad \cup \left\{\mathbf{x} \in E \setminus (E_1 \cup E_2) \ \big| \ \left(f|_{E \setminus \{f \pm \infty\}}(\mathbf{x}) \right) \cdot \left(g|_{E \setminus \{g \pm \infty\}}(\mathbf{x}) \right) > a \right\},
\end{align}which is also measurable. This proves that h is measurable. Therefore, fg is measurable. 
Now, let f, g: E \to \overline{\R} be measurable. Given a \in \R, define f + g = a whenever it is of the form +\infty + (-\infty) or -\infty + \infty. Then for the set E_1 = \{f + g = \infty\}, we have E_1 = \{f \in \R, g = \infty\} \cup \{g = g = \infty\} \cup \{f = \infty, g \in \R\}which is measurable by Corollary 4.2 (from the textbook). Likewise, the set E_2 = \{f + g = - \infty\} is measurable. By Lemma 2, it suffices to show that h = (f + g)|{E \setminus (E_1 \cup E_2)} is measurable, Let b \in \R. If b \ge a, then\begin{align}
&\{\mathbf{x} \in E \setminus (E_1 \cup E_2) \mid h(\mathbf{x}) > a\} \\
&= \left\{\mathbf{x} \in E \setminus (E_1 \cup E_2) \ \big| \ \left(f|_{E \setminus \{f \pm \infty\}}(\mathbf{x}) \right) + \left(g|_{E \setminus \{g \pm \infty\}}(\mathbf{x}) \right) > b \right\}, 
\end{align}
which is measurable By Theorem 4.9 (in the textbook). 
If b < a, then\begin{align}
&\{\mathbf{x} \in E \setminus (E_1 \cup E_2) \mid h(\mathbf{x}) > b\} \\
&= \{f = \infty, g = -\infty\} \cup \{f = -\infty, g = \infty\} \\
&\quad \cup \left\{\mathbf{x} \in E \setminus (E_1 \cup E_2) \ \big| \ \left(f|_{E \setminus \{f \pm \infty\}}(\mathbf{x}) \right) + \left(g|_{E \setminus \{g \pm \infty\}}(\mathbf{x}) \right) > b \right\}
\end{align}
which is also measurable. This proves that h is measurable. Therefore, f + g is measurable.