Measurability of a measurable function times a negative constant

Question

Given a measurable space $(X,\scr{A})$ where $\scr{A}$ is a sigma algebra on $X$, I would like to know if $f:X \to \mathbb{R}$ is measurable then $cf$ where $c \in \mathbb{R}$ is also measurable.

A function $f:X \to \mathbb{R}$ is measurable if $[f > \alpha] = \{x \in X \:{:}\: f(x) > \alpha \} \in \mathscr{A}$ for all $\alpha \in \mathbb{R}$.

If $c = 0$, then $[cf > \alpha] = X$ for all $\alpha < 0$ and $[cf > \alpha] = \emptyset$ for all $\alpha \geq 0$, so $[cf > \alpha] \in \scr{A}$.

If $c > 0$, then $[cf > \alpha] = [f > \frac{\alpha}{c}] \in \scr{A}$.

My question is when $c < 0$, my attempt is as follows.

Since $[cf > \alpha] = [f < \frac{\alpha}{c}] = \{x \in X \:{:}\: f(x) < \frac{\alpha}{c} \}$ and $[f > \frac{\alpha}{c}] \in \scr{A}$, the complement of ${[f > \frac{\alpha}{c}]} $ which is $[f \leq \frac{\alpha}{c}] \in \scr{A}$, but as $[f < \frac{\alpha}{c}] \neq [f \leq \frac{\alpha}{c}]$, I am stuck in showing $[f < \frac{\alpha}{c}] \in \scr{A}$.

Any help will be greatly appreciated.

Answer 1

Since $\mathcal A$ is a $\sigma$-\algebra, asserting that$$\left\{x\in X\,\middle|\,f(x)<\frac\alpha c\right\}\in\mathcal A$$is equivalent to asserting that its complement, which is$$\left\{x\in X\,\middle|\,f(x)\geqslant\frac\alpha c\right\},\tag1$$belongs to $\mathcal A$. And$$(1)=\bigcap_{n\in\mathbb N}\left\{x\in X\,\middle|\,f(x)>\frac\alpha c-\frac1n\right\}$$Therefore, $(1)\in\mathcal A$.

Answer 2

Note that for $b=\alpha /c$ we have $[f<b] = [f\ge b]^C$ and $$[f\ge b] = \cap_{n=1}^\infty [f >b - 1/n]\in \mathscr A.$$

($A^C$ is the complement of the set $A$)

Answer 3

In general if $(X,\mathscr A)$, $(Y,\mathscr B)$ and $(Z,\mathscr C)$ are measurable spaces and $f:X\to Y$ and $g:Y\to Z$ are both measurable then also $g\circ f:X\to Z$ is measurable.

This because: $$(g\circ f)^{-1}(\mathscr C)=f^{-1}(g^{-1}(\mathscr C))\subseteq f^{-1}(\mathscr B)\subseteq\mathscr A$$

The first $\subseteq$ because $g$ is measurable and the second $\subseteq$ because $f$ is measurable.

This can be applied here for $(Y,\mathscr B)=(\mathbb R,\mathcal B)=(Z,\mathscr C)$ where $cf:X\to\mathbb R$ can be recognized as $g\circ f$ if the function $g:\mathbb R\to\mathbb R$ is prescribed by $x\mapsto cx$ where $c\in\mathbb R$. It is evident that this function $g$ is measurable.