Measurable Functions Determine Measurable Sets

Theorem
Let $\left({X, \Sigma}\right)$ be a measurable space.

Let $f, g: X \to \overline{\R}$ be $\Sigma$-measurable functions.

Then the following sets are measurable:


 * $\left\{{f < g}\right\}$
 * $\left\{{f \le g}\right\}$
 * $\left\{{f = g}\right\}$
 * $\left\{{f \ne g}\right\}$

where, for example, $\left\{{f < g}\right\}$ is short for $\left\{{x \in X: f \left({x}\right) < g \left({x}\right)}\right\}$.

Proof
From Pointwise Difference of Measurable Functions is Measurable, $f - g: X \to \overline{\R}$ is $\Sigma$-measurable.

Now we have the following evident identities:


 * $\left\{{f < g}\right\} = \left\{{f - g < 0}\right\}$
 * $\left\{{f \ge g}\right\} = \left\{{f - g \le 0}\right\}$
 * $\left\{{f = g}\right\} = \left\{{f - g = 0}\right\}$
 * $\left\{{f \ne g}\right\} = \left\{{f - g \ne 0}\right\}$

Subsequently, using the preimage under $f - g$, the latter may respectively be expressed as:


 * $\left\{{f - g < 0}\right\} = \left({f - g}\right)^{-1} \left({\left({-\infty \,.\,.\, 0}\right)}\right)$
 * $\left\{{f - g \ge 0}\right\} = \left({f - g}\right)^{-1} \left({\left({-\infty \,.\,.\, 0}\right]}\right)$
 * $\left\{{f - g = 0}\right\} = \left({f - g}\right)^{-1} \left({\left\{{0}\right\}}\right)$
 * $\left\{{f - g \ne 0}\right\} = \left({f - g}\right)^{-1} \left({\R \setminus \left\{{0}\right\}}\right)$

Now the sets:


 * $\left({-\infty \,.\,.\, 0}\right)$
 * $\left({-\infty \,.\,.\, 0}\right]$
 * $\left\{{0}\right\}$
 * $\R \setminus \left\{{0}\right\}$

are all open or closed in the Euclidean topology.

Hence by definition of Borel $\sigma$-algebra, and by Closed Set Measurable in Borel Sigma-Algebra, they are $\mathcal B \left({\R}\right)$-measurable.

Since $f - g$ is $\Sigma$-measurable, it follows that:


 * $\left\{{f < g}\right\}$
 * $\left\{{f \le g}\right\}$
 * $\left\{{f = g}\right\}$
 * $\left\{{f \ne g}\right\}$

are all $\Sigma$-measurable sets.