Finding a function $\phi$ so that $\phi(X)$ is a random variable In my notes, Chebyshev's inequality is formulated as follows:
$$
\forall a \geq 0: P(X \geq a) \leq \dfrac{1}{\phi(a)}E[\phi(X)]
$$
Which is true when $X$ is a random variable and  $\phi\geq 0$ and non-decreasing for $-\infty &lt; x &lt; \infty$.

$\phi(X)$ is simply said to be a function $\phi: \mathbb{R} \rightarrow \mathbb{R}$, so that X is a random variable.

Which conditions apply so that $\phi(X)$ is still a random variable?
Edited answer after realizing my mistake:

The function $\phi$ only needs to be measurable for $\phi(X)$ to be a random variable. As @Did says, a non-decreasing function is trivially measurable.

More formally, if $X$ is a random variable defined on  $(\Omega,\mathscr A)$, then for any Borel (measurable) function $\phi$ on $\mathbb R$, $\phi(X)$ is also a random variable.

This is because for all $x\in\mathbb R$, we have $$\{\omega\in\Omega:\phi(X(\omega))\le x\}=\{\omega\in\Omega:X(\omega)\in \phi^{-1}((-\infty,x])\}$$ 

As $\phi$ is measurable, $\phi^{-1}((-\infty,x])$ is a Borel (measurable) set and as $X$ is a random variable, $$\{\omega\in\Omega:X(\omega)\in \phi^{-1}((-\infty,x])\}\in \mathscr A\quad,\forall\,x\in\mathbb R$$
Edited answer after realizing my mistake:

The function $\phi$ only needs to be measurable for $\phi(X)$ to be a random variable. As @Did says, a non-decreasing function is trivially measurable.

More formally, if $X$ is a random variable defined on  $(\Omega,\mathscr A)$, then for any Borel (measurable) function $\phi$ on $\mathbb R$, $\phi(X)$ is also a random variable.

This is because for all $x\in\mathbb R$, we have $$\{\omega\in\Omega:\phi(X(\omega))\le x\}=\{\omega\in\Omega:X(\omega)\in \phi^{-1}((-\infty,x])\}$$ 

As $\phi$ is measurable, $\phi^{-1}((-\infty,x])$ is a Borel (measurable) set and as $X$ is a random variable, $$\{\omega\in\Omega:X(\omega)\in \phi^{-1}((-\infty,x])\}\in \mathscr A\quad,\forall\,x\in\mathbb R$$