# Measure theory: measure zero

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 $$c then there is some rational $$q$$ with $$c\le q. 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 $$\mathbb{Q}$$?
• 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? – StMan Oct 21 '18 at 19:25