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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.

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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 $\mathbb{Q}$?

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  • $\begingroup$ 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

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