Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

(1) Suppose a function $f$ has a [Lebesgue] measurable domain and is continuous except at a finite number of points. Is $f$ is necessarily [Lebesgue] measurable?
Comments For (1), If $f$ is defined on a [Lebesgue] measurable set $E$ and is continuous except for a finite number of values say $x_1,x_2, ... x_n$ are those points of discontinuity, then can't we describe the pre-images of $f$ just as a finite union of of the pre-images of the collection of continuous functions $\{f_i\}_{i=1}^{n}$, where each $f_i$ is defined up between each point of discontinuity of $f$? Or am I missing something here?

(2) Suppose the function $f$ is defined on a measurable set $E$ and has the property that $\{x \in E | f(x) > c\}$ is measurable for each rational number $c$. Is $f$ necessarily [Lebesgue] measurable?
Comments I got this one, thanks to everyone who commented.

Any hints would be appreciated. The text being used is Royden-Fitzpatrick 4th Edition.

share|improve this question
2  
$\{x \in E | f(x) > c\}=\cup_{q>c, q\in\mathbb{Q}}\{x \in E | f(x) > q\}$ is a countable union of measurable sets –  yoyo May 11 '13 at 18:45
    
Gah, I feel dumb. Thanks –  Archie May 11 '13 at 23:31
    
Also, I edited what I have observed for (1), but I'm still not quite there yet. –  Archie May 11 '13 at 23:32
    
@Archie If you look my answer I also answered the question assuming Lebesgue measurable functions. –  Gastón Burrull May 12 '13 at 22:40

2 Answers 2

(2) It is true, this not contradicts the proposition 1 (Page 54), just is an equivalent form of proposition 1.

Suppose $\{x\in E : f(x)>c\}$ measurable for each rational $c$. Let $r\in\mathbb{R}$, then exist a sequence $\{c_n\}$ of rational numbers such that $c_n\to r$ and $c_n\geq r$ for all $n\in\mathbb{N}$. Note that

$$A=\{x\in E : f(x)>r\}=\{x\in E : f(x)\leq r\}^c=\left(\bigcap\{x\in E : f(x)\leq c_n\}\right)^c=\bigcup \{x\in E : f(x)> c_n\}$$

Hence $A$ is measurable. Try (1) by yourself.

(1) When you say "where each $f_i$ is defined up between each point of discontinuity of $f$" you are wrong, $E$ does not necessarily has a good order, we only know that $E$ is a topological space. If you don't assume that each open set in a topology space must be measurable, (1) is obviously false (indeed, proposition 3 in page 55 Royden 4th ed. will fail). The correct word in this Royden Chapter is Lebesgue Measurable functions instead of measurable function, then we know that $E\subset\mathbb{R}$.

If $E\subset\mathbb{R}$ (1) is true. Instead of considering $f_n$ consider simply $f|\{E\setminus{x_1,\ldots,x_n}\}$, define $f(x_i)=y_i\in\mathbb{R}$. What happen when an open interval $I$ contains some of $y_i$? (in this case is much easier work with $y_n$ rather than $x_n$)

How fails "preimage of open set is open" when a continuous function $f\colon\mathbb{R}\to\mathbb{R}$ has finitely many discontinuities?. Note that the only case of preimage of an open interval that is not an open set correspond to closed intervals, semi-open intervals (for example in $[a,b)$ form) or numerable union of them, so that not only the function $f$ is measurable, it is also Borel measurable (preimage of open sets are Borel Measurable).

In any topological space $(X,\tau)$ where each open set is measurable, and consider as measurable sets the Borel $\sigma$-algebra (the smallest sigma algebra containing the open sets) I don't know if (1) is true.

share|improve this answer
    
I made an edit for (1), am I on the right track? I'm just having issues sketching the details. –  Archie May 12 '13 at 0:00
    
@Archie I edited my answer with (1). –  Gastón Burrull May 12 '13 at 2:26
    
I guess I should have stated Lebesgue measurable set $E$ and we working under the standard topology. But thank you for the input. –  Archie May 12 '13 at 17:00
    
However when we define $f(x_i) = y_i$ wouldn't there be cases when this would be undefined since we stated that the $x_i$'s are the points in which $f$ is not continuous? –  Archie May 12 '13 at 17:02
1  
@Archie $\frac{1}{x}$ is not discontinuous at $0$ is undefined in $0$, that means the domain does not contains $0$, then $f$ is continuous in the whole domain $\mathbb{R}\setminus\{0\}$. If you define in cases $f$ by $f(x)=877$ if $x=0$ and $f=\frac{1}{x}$ if $x\neq 0$ then $f$ has only one discontinuity in $0$ (in this case $x_1=0$, $y_1=877$). –  Gastón Burrull May 12 '13 at 21:40

What about this for the first question:

Let $A$ be the measurable domain of $f$ and let $X=\{x_i\}_{i=1}^n$ be the finite collection of discontinuities. Since $|X|$ is finite, we can order them from smallest, say $x_1$, to largest, say $x_n$. Then for $1\le i\le n+1$ define $$ A_i=A\cap (x_{i-1},x_i), $$ where $x_0=-\infty$ and $x_{n+1}=\infty$. Observe that each $A_i$ is measurable and $$ \{x\in A\,:\, f(x)>c\}=\left(\bigcup_{i=1}^{n+1}\{x\in A_i\,:\, f(x)>c\}\right) \cup \{x\in X\,:\,f(x)>c\} $$ For each $c$, each set in the indexed union is measurable (because $f$ is measurable on each A_i) and each subset of $X$ is either empty or contains no more than $n$ points and is therefore measurable as well.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.