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

• MHB
• Math Amateur
In summary: X | (f+g) \lt r \} = \{ x \in X | \ (f+g) \lt r \} \subseteq \{ x \in X | f(x) \lt q \} \cap \{ x \in X | g(x) \lt r - q \}$Math Amateur Gold Member MHB 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=============================================================================================================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 Have you tried proving mutual inclusion? Hi Evgeny ... ... can you please expand on what you mean ... I’m a bit lost ... Peter Don't you know that$A=B\iff A\subseteq B\land B\subseteq A$and$X\subseteq Y\iff\forall x.\,x\in X\implies x\in Y$? Oh sorry ... yes, indeed, understand that... ... will try following your suggestion... Thanks for the help ... Peter 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: Peter said: 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$q$. 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 Peter said: 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.)

## 1. What is a measurable function?

A measurable function is a function that maps elements from one measurable space to another measurable space, where measurable spaces are sets equipped with a sigma-algebra. Measurable functions are important in the study of measure theory and are used to define integrals and probability measures.

## 2. What is the significance of Lindstrom's Proposition 7.3.7 in measurable functions?

Lindstrom's Proposition 7.3.7 states that any measurable function between two measurable spaces can be approximated by a sequence of simple functions. This result is significant because it allows us to approximate complex measurable functions with simpler ones, making it easier to calculate integrals and probabilities.

## 3. Can you provide an example of a measurable function?

One example of a measurable function is the indicator function, which maps a set to either 0 or 1 depending on whether an element is in the set or not. This function is measurable because it maps elements from a measurable space (the set) to a measurable space (the set of real numbers).

## 4. How is the concept of measurable functions related to measure theory?

Measurable functions are closely related to measure theory because they are used to define integrals and probability measures. In measure theory, a measure is a function that assigns a non-negative value to sets, and measurable functions are used to calculate the measure of a set.

## 5. Are all continuous functions also measurable functions?

No, not all continuous functions are measurable. A function is measurable if the pre-image of any measurable set is a measurable set. While all continuous functions are Borel-measurable (meaning the pre-image of any Borel set is a Borel set), there are measurable functions that are not continuous.

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