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Below is a proof to show that if $f$ is a real function on a measurable space $X$ such that $\{x : f(x) \gt r\}$ is measurable for every irrational $r$, then $f$ is measurable.

Suppose $\{x \in X | f(x) \gt r\}$ is measurable, $r\in \mathbb{R\setminus Q}$. Let $d\in\mathbb{R}$. Then for each $n$ $\exists$ $r_n\in\mathbb{R\setminus Q}$ such that $r_n \gt d$. Then $$ \{x | f(x) \gt d\} = \bigcup_{n=1}^\infty \{x | f(x) \gt r_n\}.$$ Since $\{x | f(x) \gt r_n\} $ is measurable, $f$ is measurable.

Please, is this right?

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    Why are the $\{x|f(x)>r_n\}$ measurable even though the $r_n$ are rational? Your hypothesis allows this only for irrational values,no?2012-05-02
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    @FortuonPaendrag: it was a typo. I fixed it. does it look okay?2012-05-02
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    You need one final condition. The $r_n$s must converge to $d$. Do you agree?2012-05-02
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    @FortuonPaendrag: why must they converge to $d$?2012-05-02
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    Because, for equality to hold in the equation with the countable union, you must be able to bring the $r_n$s as close to $d$ as possible. Otherwise you risk not treating some values in between. As an example, if my $d=3$, if I chose my $r_1=\pi,r_2=2\pi \cdots$, this is not enough, because I left out the values in the interval $(3,\pi)$. That is, if for some $x$, $f(x)=3.1$, then $x$ will be in the set to the left, but not the right.2012-05-02
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    @Linda Well if the sequence you have chosen doesn't converge to $d$, then they could go "away" from $d$ and your union equation wouldn't work out, right? If you make them converge to $d$, then the union eventually fills up the LHS of your equation.2012-05-02
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    Thanks to you both for the explanation.2012-05-02
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    Do let me know if you would like me to elaborate further. It would be a pleasure!2012-05-02
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    @FortuonPaendrag: please, if you don't mind...thank you. Perhaps, you could post your summations as an answer.2012-05-02

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My problem is with this statement in your proof. The rest seems fine.

$$\displaystyle \{x|f(x)>d\}=\bigcup_{n=1}^\infty \{x|f(x)>r_n\}.$$

If you do not place suitable restrictions upon your $r_n$, I do not think it is true.

For example, see the case when $d=3$. We will see that for some $r_n$s not chosen carefully, that equality is not true. Choose $r_n=n\pi$. It is a sequence of irrational numbers, and they are all bigger than $d=3$. Just like your condition.

BUT, if there was an $a \in X$ for which $f(a)$ was say $3.1$ (just to give an example), then, since $3.1$ is bigger than $3$, $a \in \{x|f(x)>3\}$, the LHS . But since $3.1$ is smaller than any $n\pi$ for positive $n$, $a\notin \displaystyle \bigcup_{n=1}^\infty \{x|f(x)>r_n\}$. So those sets are not equal.

To remedy this, we must eliminate that "gap" we saw between $3.1$(or I could have done all of this with 3.01 or 3.001 or as close to 3 as I wanted) and the $r_n$s. So, the way to make sure this doesnt happen, is to make sure that $r_n \to d$.

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    @Linda Do not hesitate to ask for clarifications! This is a wonderful topic and you should have your doubts cleared to enjoy it fully!2012-05-02
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With $d \in \mathbb{Q}$, choose irrationals $r_n$ such that $r_n \downarrow d$ (ie, each $r_n \geq d$, and $r_n$ converges to $d$). Then your construction works, because the set on the right is the countable union of measurable sets.