Proof of the fact that $\{ x \in \mathbb{R^n} | f(x)\neq g(x) \}$ is lebesgue-measurable.

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Let $f,g: \mathbb{R^n} \to [-\infty, \infty]$ be measurable functions.

Then, prove that $A:=\{ x \in \mathbb{R^n} \ | \ f(x)\neq g(x) \}$ is a lebesgue-measurable set.

Is my proof correct?

My proof is here.

$A$ can be written as $A=\{x \in \mathbb{R^n} \ | \ f(x)> g(x) \} \cup \{x \in \mathbb{R^n} \ | \ f(x)< g(x) \}$.

Let $B=\{x \in \mathbb{R^n} \ | \ f(x)> g(x) \} $, $C=\{x \in \mathbb{R^n} \ | \ f(x)< g(x) \}$.

Then, $B=\cup_{r\in \mathbb{Q}}\big(\{f > r \} \cap \{ g < r\}\big)$, since $B\supset \cup_{r\in \mathbb{Q}}\big(\{f > r \} \cap \{ g < r\}\big)$ is trivial, and if $x\in B,$ $f(x)>g(x)$ holds and there exists $r\in \mathbb{Q}$ s.t. $f(x)>r>g(x)$ thus $x \in \{f>r\} \cap \{g< r\}\subset \cup_{r\in \mathbb{Q}}\big(\{f > r \} \cap \{ g < r\}\big).$

And now, since $f$ and $g$ are measurable, $\{ f>r \}$ and $\{ g<r \}$ are measurable and therefore $B$ is measurable.

In the same way, we can see that $C$ is also mesurable.

Therefore, $A=B\cup C$ is measurable.