# Measurable Functions ... Lindstrom, Proposition 7.3.7 ... ...

#### Peter

##### Well-known member
MHB Site Helper
I am reading Tom L. Lindstrom's book: Spaces: An Introduction to Real Analysis ... and I am focused on Chapter 7: Measure and Integration ...

I need help with the proof of Proposition 7.3.7 ...

Proposition 7.3.7 and its proof read as follows:

In the above proof by Lindstrom we read the following:

" ... ... $$\displaystyle (f + g)^{ -1} ( [ - \infty , r ) ) = \{ x \in X | (f + g) \lt r \}$$

$$\displaystyle = \bigcup_{ q \in \mathbb{Q} } ( \{ x \in X | f(x) \lt q \} \cap \{ x \in X | g \lt r - q \} )$$ ... ... "

Can someone please demonstrate, formally and rigorously, how/why ...

$$\displaystyle \{ x \in X | (f + g) \lt r \} = \bigcup_{ q \in \mathbb{Q} } ( \{ x \in X | f(x) \lt q \} \cap \{ x \in X | g \lt r - q \} )$$ ... ...

Help will be much appreciated ...

Peter

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Readers of the above post may be assisted by access to Lindstrom's introduction to measurable functions (especially Lindstrom's definition of a measurable function, Definition 7.3.1) ... so I am providing access to the relevant text ... as follows:

Hope that helps ...

Peter

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Have you tried proving mutual inclusion?

#### Peter

##### Well-known member
MHB Site Helper
Hi Evgeny ...

... can you please expand on what you mean ... I’m a bit lost ...

Peter

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
Don't you know that $A=B\iff A\subseteq B\land B\subseteq A$ and $X\subseteq Y\iff\forall x.\,x\in X\implies x\in Y$?

#### Peter

##### Well-known member
MHB Site Helper
Oh sorry ... yes, indeed, understand that...

... will try following your suggestion...

Thanks for the help ...

Peter

#### Peter

##### Well-known member
MHB Site Helper
I have been reflecting on proving that $$\displaystyle f+ g$$ is measurable when $$\displaystyle f$$ and $$\displaystyle g$$ are measurable ... and I need to clarify an issue before trying the mutual inclusion argument ...

I need to understand exactly what is wrong with the following argument ... something surely is wrong because the proof doesn't involve ranging over $$\displaystyle \mathbb{Q}$$ ...

... ... ... the argument follows ... ...

First note that $$\displaystyle f + g \lt r$$

$$\displaystyle \Longrightarrow$$ there exists $$\displaystyle q \in \mathbb{Q}$$ such that $$\displaystyle f \lt q \lt r - g$$

$$\displaystyle \Longrightarrow f \lt q$$ and $$\displaystyle g \lt r - q$$

$$\displaystyle \Longrightarrow$$ for all $$\displaystyle x \in X$$ we have $$\displaystyle f(x) \lt q$$ and $$\displaystyle g(x) \lt r - q$$

Thus $$\displaystyle \{ x \in X \ | \ (f + g) \lt r \} = \{ x \in X \ | \ f(x) \lt q \} \cap \{ x \in X \ | \ g(x) \lt r - q \}$$

Can someone please clarify what is wrong with the above argument ...

Help will be much appreciated ...

Peter

Last edited:

#### Opalg

##### MHB Oldtimer
Staff member
I have been reflecting on proving that $$\displaystyle f+ g$$ is measurable when $$\displaystyle f$$ and $$\displaystyle g$$ are measurable ... and I need to clarify an issue before trying the mutual inclusion argument ...

I need to understand exactly what is wrong with the following argument ... something surely is wrong because the proof doesn't involve ranging over $$\displaystyle \mathbb{Q}$$ ...

... ... ... the argument follows ... ...

First note that $$\displaystyle f + g \lt r$$

$$\displaystyle \Longrightarrow$$ there exists $$\displaystyle q \in \mathbb{Q}$$ such that $$\displaystyle f \lt q \lt r - g$$

$$\displaystyle \Longrightarrow f \lt q$$ and $$\displaystyle g \lt r - q$$

$$\displaystyle \Longrightarrow$$ for all $$\displaystyle x \in X$$ we have $$\displaystyle f(x) \lt q$$ and $$\displaystyle g(x) \lt r - q$$

Thus $$\displaystyle \{ x \in X \ | \ (f + g) \lt r \} = \{ x \in X \ | \ f(x) \lt q \} \cap \{ x \in X \ | \ g(x) \lt r - q \}$$

Can someone please clarify what is wrong with the above argument ...

Help will be much appreciated ...

Peter
Your argument correctly shows that if $(f+g)(x)<r$ then there exists $q \in \mathbb{Q}$ such that $$\displaystyle f(x) \lt q$$ and $$\displaystyle g(x) \lt r - q$$. But the choice of $q$ depends on $x$. If you want to eliminate that dependency you then have to take the union of all $q\in\Bbb{Q}$, to get $$\displaystyle \{ x \in X | (f + g) \lt r \} \subseteq \bigcup_{ q \in \mathbb{Q} } ( \{ x \in X | f(x) \lt q \} \cap \{ x \in X | g \lt r - q \} )$$.

Lindstrom conceals this difficulty by using the horrible notation $f+g<r$ rather than $(f+g)(x)<r$, which disguises the fact that as $x$ varies so does $q$.

#### Peter

##### Well-known member
MHB Site Helper
Thanks for the help Opalg ...

I can see $$\displaystyle q$$ may depend on $$\displaystyle x$$ ... but cannot see why a union over all $$\displaystyle \mathbb{Q}$$ resolves this issue ...

... why are we justified in taking a union over all $$\displaystyle \mathbb{Q}$$ ...

Can you explain further ...

Peter

#### Opalg

##### MHB Oldtimer
Staff member
Thanks for the help Opalg ...

I can see $$\displaystyle q$$ may depend on $$\displaystyle x$$ ... but cannot see why a union over all $$\displaystyle \mathbb{Q}$$ resolves this issue ...

... why are we justified in taking a union over all $$\displaystyle \mathbb{Q}$$ ...

Can you explain further ...

Peter
I'll try to make it clearer by putting a subscript $0$ for a particular point $x_0$ and a particular rational number $q_0$.

You have shown that, given $x_0\in X$ satisfying $(f+g)(x_0)<r$, there exists $q_0\in\Bbb{Q}$ such that $$x_0 \in \{ x \in X | f(x) \lt q_0 \} \cap \{ x \in X | g \lt r - q_0 \} \subseteq \bigcup_{ q \in \mathbb{Q} } ( \{ x \in X | f(x) \lt q \} \cap \{ x \in X | g \lt r - q \} ).$$ Since that holds for all $x_0$ satisfying $(f+g)(x_0)<r$, it follows that $$\{x\in X | (f+g)(x)<r\} \subseteq \bigcup_{ q \in \mathbb{Q} } ( \{ x \in X | f(x) \lt q \} \cap \{ x \in X | g \lt r - q \} ).$$

(You still have to prove the reverse inclusion, as Evgeny suggests. But that should not be difficult.)