I'm studying real analysis. Below is my attempt at solving problem 2 in Ch 2.1 Folland. Could someone please tell me if they see any mistakes. Thanks in advance.
Problem: Suppose $f,g:X \rightarrow \bar{\mathbb{R}}=[-\infty, \infty]$ are $M-$measurable. Prove that
a. $fg$ is measurable.
b. If $a \in \bar{\mathbb{R}}$ is fixed, $h(x)=a$ if $f(x)=-g(x)=\pm \infty$ and $h(x)=f(x) + g(x)$ otherwise, then $h$ is measurable.
Proof: a. It suffices to show that $(fg)^{-1}(\{\pm\infty\}) \in M$ and that $fg$is measurable on $Y:=(fg)^{-1}(\mathbb{R}).$ We can write $(fg)^{-1}(\{\infty\})=f^{-1}(\{\infty\}) \cap g^{-1}((0, \infty]) \cup g^{-1}(\{\infty\}) \cap f^{-1}((0, \infty]) \cup f^{-1}(\{-\infty\}) \cap g^{-1}([-\infty,0)) \cup g^{-1}(\{-\infty\}) \cap f^{-1}([-\infty,0)) \mbox{ and }(fg)^{-1}(\{-\infty\})=f^{-1}(\{-\infty\}) \cap g^{-1}((0, \infty]) \cup g^{-1}(\{-\infty\}) \cap f^{-1}((0, \infty]) \cup f^{-1}(\{\infty\}) \cap g^{-1}([-\infty,0)) \cup g^{-1}(\{\infty\}) \cap f^{-1}([-\infty,0)).$ Since $f$ and $g$ are measurable, it follows that $(fg)^{-1}(\{\infty\}), (fg)^{-1}(\{-\infty\}) \in M.$
Let $B \in B_{\bar{\mathbb{R}}}$ and note that $(fg)^{-1}(B) \cap Y = (fg)^{-1} (B \cap \mathbb{R}),$ where $B \cap \mathbb{R} \in B_{\mathbb{R}}.$ Define $X'=X \backslash ((fg)^{-1}(\{-\infty\}) \cup (fg)^{-1}(\{\infty\}))$ and $fg'=fg|_{X'}:X' \rightarrow \mathbb{R}.$ Let $F:X'\rightarrow \mathbb{R}^2$ be given by $F(x)=(f(x), g(x)),$ and $\psi: \mathbb{R}^2 \rightarrow \mathbb{R}$ by $\psi(z,w)=zw.$ Then $F$ is $(M, B_{\mathbb{R}} \otimes B_{\mathbb{R}})-$measurable because $f$ and $g$ are $(M, B_{\mathbb{R}})-$measurable and $\psi$ is $(B_{\mathbb{R}} \otimes B_{\mathbb{R}}, B_{\mathbb{R}})-$measurable because it is continuous, so $fg'=\psi \circ F$ is $(M, B_{\mathbb{R}})-$measurable. For any $B' \in B_{\mathbb{R}},$ $fg'^{-1}(B')=(fg)^{-1}(B').$ So $(fg)^{-1}(B') \in M$ and in particular $(fg)^{-1}(B \cap \mathbb{R}) \in M,$ so $fg$ is measurable on $Y.$
b. Let $B \in B_{\bar{\mathbb{R}}}.$ Based on the definition of $h,$ we can write $h^{-1}(B)=(f+g)^{-1}(B \backslash \{a\}) \cup (f+g)^{-1}(\{0\}) \cap g^{-1}(\{\infty\}) \cup (f+g)^{-1}(\{0\}) \cap g^{-1}(\{-\infty\}).$ Since $g$ is measurable, $g^{-1}(\{-\infty\}), g^{-1}(\{\infty\}) \in M.$ Now put $X'=X \backslash (f^{-1}(\{\pm \infty\}) \cup g^{-1}(\{\pm \infty\}))$ and define $f+g'=(f+g)|_{X'}:X' \rightarrow \mathbb{R}.$ As before, take $F:X' \rightarrow \mathbb{R}^2$ to be $F(x)=(f(x), g(x))$ and $\phi: \mathbb{R}^2 \rightarrow \mathbb{R}$ to be $\phi(z,w)=z+w.$ Then, $f+g'=\phi \circ F$ is $(M, B_{\mathbb{R}})-$measurable, $(f+g)^{-1}(\{0\}) \in M$ and $(f+g)^{-1}((B \backslash \{a\}) \backslash \{-\infty, \infty\}) \in M.$ Once we show that $(f+g)^{-1}((B \backslash \{a\}) \cap \{-\infty, \infty\}) \in M,$ it will follow that $h^{-1}(B) \in M$ and hence that $h$ is measurable. If (1) $(B \backslash \{a\}) \cap \{-\infty, \infty\}=\{\infty\},$ then $(f+g)^{-1} (\{\infty\})= f^{-1}(\{\infty\}) \cap g^{-1} ((-\infty, \infty]) \cup g^{-1}(\{\infty\}) \cap f^{-1}((-\infty, \infty] \in M,$ if (2) $(B \backslash \{a\}) \cap \{-\infty, \infty\}=\{-\infty\},$ then $(f+g)^{-1} (\{-\infty\})= f^{-1}(\{-\infty\}) \cap g^{-1} ([-\infty, \infty)) \cup g^{-1}(\{-\infty\}) \cap f^{-1}([-\infty, \infty) \in M,$ and if $(B \backslash \{a\}) \cap \{-\infty, \infty\}=\{-\infty, \infty\},$ then by (1) and (2), $(f+g)^{-1} (\{\pm \infty\})=(f+g)^{-1} (\{-\infty\}) \cup (f+g)^{-1} (\{\infty\}) \in M.$