# Measure Theory “Magical” Unions and Intersections

In reading Knapp's Basic Real Analysis, in chapter 5 section 2, I came across something which I just cannot understand. If $f$ and $g$ are measurable functions, then he says that $$(f+g)^{-1}(c,+\infty)=\bigcup_{r\in \mathbb{Q}}f^{-1}(c+r,+\infty]\cap g^{-1}(-r,+\infty].$$ I spent a lot of time thinking and working with how he derived this, or even how the conclusion holds, but I cannot. So, if someone could help me figure out exactly how this is true, and how this "magical" equation was derived, then I would be very appreciative.

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You should say $r\in\mathbb{Q}$. Does this help? – ncmathsadist Jan 28 '13 at 0:17
That was just a typo, but it still doesn't address my lack of understanding unfortunately. – Jebruho Jan 28 '13 at 0:19

Suppose that $x\in(f+g)^{-1}\big[(c,\to)\big]$; then $(f+g)(x)=f(x)+g(x)>c$. Let $s=f(x)-c$; $f(x)=c+s$, and therefore $g(x)>-s$. We’d like to find a rational $r$ such that $f(x)>c+r$ and $g(x)>-r$, since we would then have $$x\in f^{-1}\big[(c+r,\to)\big]\cap g^{-1}\big[(-r,\to)\big]\;.\tag{1}$$ Since $f(x)=c+s$, this implies that we want $c+s>c+r$, or $r<s$. We also need to have $g(x)>-r$. In short, we need a rational number $r$ such that $-s<-r<g(x)$, and since $-s<g(x)$ and the rationals are dense in $\Bbb R$, there must be one.

We’ve now shown that for each $x\in(f+g)^{-1}\big[(c,\to)\big]$ there is an $r\in\Bbb Q$ such that $(1)$ holds, and it follows at once that

$$(f+g)^{-1}\big[(c,\to)\big]\subseteq\bigcup_{r\in\Bbb Q}\left(f^{-1}\big[(c+r,\to)\big]\cup g^{-1}\big[(-r,\to)\big]\right)\;.$$

The opposite inclusion is easier. If

$$x\in\bigcup_{r\in\Bbb Q}\left(f^{-1}\big[(c+r,\to)\big]\cup g^{-1}\big[(-r,\to)\big]\right)\;,$$, there is an $r\in\Bbb Q$ such that $(1)$ holds. It follows that $f(x)>c+r$ and $g(x)>-r$ and hence that $(f+g)(x)=f(x)+g(x)>c+r-r=c$, i.e., that $x\in(f+g)^{-1}\big[(c,\to)\big]$.

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Great answer! Thank you. – Jebruho Jan 28 '13 at 0:55
@Jebruho: You’re welcome (and thank you). – Brian M. Scott Jan 28 '13 at 0:57

It might help to consider the following equivalent statement: for any $x$, we have $f(x) +g(x) > c$ iff there exists $r \in \mathbb{Q}$ such that $f(x) > c+r$ and $g(x) > -r$.

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Suppose that $f(x) + g(x) < c$. Then since $f(x) < c - g(x)$, you can choose a rational number so that $f(x) < r < c - g(x)$. You have $x\in[f < r]$. Now $c - g(x) > r$, so $g(x) < c - r$; therefore $x \in [g < c - r]$. We have just shown that $$[f + g < c] = \bigcup_{r\in \mathbb{Q} }[f < r]\cap [ g < c - r].$$

The reverse inequality is easier.

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