Let $(X, \mathbb A)$ be a measurable space on which a measure $m$ is defined.

If $f,g : (X, \mathbb A) \to (\mathbb R, B (\mathbb R))$ are such that for any $a \in \mathbb R$ it holds that $m(\{x \in X: f(x) \leq a < g(x)\}) = 0$, then $m(\{x \in X: f(x) < g(x)\}) = 0$

How can I perfectly show this?

Intuitively, it is clear since I can choose any $a$ which is greater or equal f(x), but less than g(x). If this measure then is zero, I can also vanish the $a$ - f(x) is still less than g(x) and this has still measure 0.


Remember that $\mathbb{Q}$ is countable and dense (in $\mathbb{R}$; when dealing with $\mathbb{R}^n$, you could use $\mathbb{Q}^n$, the set of points with all-rational coordinates).

Why is this useful? Well, by density if $c<d$ then there is some rational $q$ with $c\le q<d$. You're interested in situations where $c=f(x)$ and $d=g(x)$.

Countability then comes in when we use subadditivity - specifically, the fact that the union of countably many measure zero sets is measure zero. Do you see how to decompose your set $\{x\in X: f(x)< g(x)\}$ into countably many measure-zero pieces, using the nice properties of $\mathbb{Q}$?

  • $\begingroup$ Ok, one could say that $f(x) = \frac{1}{n} < \frac{1}{n-1} < ... < \frac{1}{1} = g(x) $. Do you mean it like this? $\endgroup$ – StMan Oct 21 '18 at 19:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.