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I remember reading an example in a textbook that went something like this: if we take a function $\ell(f) = \int_{0}^{1}f(t)\, dt$, (with this being the Riemann integral) defined only on the set of continuous functions on $[0,1]$, then we may extend it to all bounded functions in $[0,1]$ by noting that $p(f) = \sup_{[0,1]}(|f(x)|)$ satisfies the criteria of the Hahn-Banach theorem.

The textbook went on to note that this defined a finitely additive set function on every subset of $[0,1]$ via allowing our set function $w(A) = \ell(\chi_{A})$. It goes on to note that, for Borel sets, this actually forms the Lebesgue measure.

Question 1. What is the reason for the "finitely additive" part of $w$ in the latter paragraph? Is it just because we required bounded functions?

Question 2. How can one see that this forms the Lebesgue measure if we consider Borel Measurable sets? I do not think this comes from Hahn-Banach.

Question 3. Is the reason that we cannot easily extend this to a larger domain (like, say, rational functions on $[0,1]$ or something) that the $\sup$ function is no longer adequate, and there is no longer a function which satisfies Hahn-Banach?

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For question 1 : It is only finitely additive, because if you have $A_1, A_2, .., A_n$ disjoint sets that have indicator functions $\chi_i$ that are Riemann-integrable, then you have $$ w\left(\bigcup_{i=1}^n A_i\right) = \ell \left( \bigcup_{i=1}^n A_i \right) = \int_0^1 \sum_{i=1}^n \,\chi_i(t) \, dt = \sum_{i=1}^n \int_0^1 \chi_i(t) \, dt = \sum_{i=1}^n w(A_i) $$ because the Riemann integral is linear. But if you take an infinite number of sets (say countable to start with), the permutation of the integral and the sum is not justified, because the move $$ w\left(\bigcup_{i=1}^{\infty} A_i\right) = \ell \left( \bigcup_{i=1}^{\infty} A_i \right) = \int_0^1 \lim_{n \to \infty} \, \sum_{i=1}^n \,\chi_i(t) \, dt = \lim_{n \to \infty} \int_0^1\sum_{i=1}^n \chi_i(t) \, dt $$ is generally justified by the fact that the sequence $\sum_{i=1}^n \chi_i(t)$ converges uniformly towards $\sum_{i=1}^{\infty} \chi_i$, and it clearly does not, thus we do not have proof. Note that if we use Lebesgue integration, this move is justified by the monotonic convergence theorem (or the dominated convergence theorem, because the $\chi_i$ are positive functions bounded by the $1$ function.).

For question 2 : You don't get the Lebesgue measure if you use the Riemann-integral to define $w$; take the indicator function of the rational numbers in $[0,1]$ as an example (the rational numbers in $[0,1]$ form indeed a Borel set) to see that you can't integrate every indicator function of any Borel set using Riemann. If you use the Lebesgue-integral to define $w$ then this is just trivial, because integrating $1$ over a measurable set just gives you its measure (this is partially how the Lebesgue integral is built, i.e. by taking functions that admit a finite number of output values).

For question 3 : I don't know if you can or cannot extend $\ell$ to a bigger set than the one of bounded functions, but the reason we don't know if we can is because we can't either prove there exists no extensions for $\ell$ at all, or because if there is an extension, we would prove there is using Hahn-Banach's theorem with a particular semi-norm that is majoring the Riemann integral. Note that this would be just one way to do it ; you can't prove things in a single manner only... although I'd be surprised to see something else come up.

EDIT : I think you might want to see this : since there are other generalizations of the Riemann integral, I thought they could be other choices for extensions of $\ell$ to different vector spaces. If you take a look at the Henstock-Kurzweil integral (let's say HK for short), I have never seen the proof of those results, but I remember reading on that page:

http://en.wikipedia.org/wiki/Henstock-Kurzweil_integral

that the space of HK integrable functions is bigger than the one of Lebesgue's functions because :

  • If $f$ is bounded, then $f$ is Lebesgue-integrable iff it is HK-integrable.
  • If $f$ is Riemann-integrable or Lebesgue-integrable, it is HK-integrable with the same value for the integral.
  • $f$ is Lebesgue-integrable if and only if $f$ and $|f|$ are HK-integrable.

Now I don't know what being HK-integrable means, but I thought you could find out yourself and maybe build some interest for this integral if it suits your problem.

Hope that helps,

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I didn't answer question 3 because I didn't understand it : Is the reason of what? You want to extend $\ell$ to something bigger than bounded functions and cannot because you used the sup function to use Hahn-Banach on the set of Riemann-integrable bounded functions and can't find another function to do so on a bigger set than the one of bounded functions? –  Patrick Da Silva Jul 23 '11 at 4:22
    
Sorry; I meant that we started with $ell$ on $C[0,1]$ and extended it to $B[0,1]$, but why couldn't we extend it to something containing $B[0,1]$. I implicitly am assuming here that we can't do it, but I'm not so sure. –  james Jul 23 '11 at 4:34
    
For question 1, I keep forgetting about the integral-sum exchange things; thank you for that answer. For question 2, I've left out a huge assumption that your counterexample reminded me to put back in. If we only assume the Borel Measurable sets in $[0,1]$, then this forms the Lebesgue measure. I'm not entirely sure why this is. –  james Jul 23 '11 at 4:38
    
I said Riemann-integrable bounded in my comment because I read your first line too quickly. Hm. Well you used Hahn-Banach by taking the semi-norm (which is actually a norm) $\sup |f|$ over the vector space of bounded functions. If you want to extend it to something bigger (or maybe different) than the set of all bounded functions, just use a different semi-norm on a bigger (or different) vector space than the one of bounded functions. –  Patrick Da Silva Jul 23 '11 at 4:40
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Oh. You're saying the extension of $\ell$ over $B[0,1]$, call it $\ell'$ for clarity, and $w(A) = \ell'(\chi_A)$, gives you the Lebesgue-measure? That is a totally different assertion, I must admit I didn't read it that way. If this is what you meant, my answer for question 2. makes no sense, I agree with you on that. The counter-example too. –  Patrick Da Silva Jul 23 '11 at 5:43

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