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Understanding Lebesgue Integration

  1. Sep 17, 2006 #1

    [tex]X[/tex] is a set.

    [tex]\mathcal{A}[/tex] is a [itex]\sigma[/itex]-algebra.

    Suppose that I have a measure space [itex](X,\mathcal{A},\mu)[/itex] and an [itex]\mathcal{A}[/itex]-measurable function:


    All pretty regular stuff. Now, I have a "supposed" measure defined as

    [tex]\nu(E):=\int_E f\mbox{d}\mu[/tex]

    for [itex]E\in \mathcal{A}[/itex]. My endeavor is to understand whether or not this function actually defines a measure on [itex]\mathcal{A}[/itex] and furthermore, understand exactly what is needed to do to work out if a given function is a measure.

    The first thing I noticed was that this function is defined using a Lebesgue integral. I think it is a Lebesgue integral because I am integrating the function over a set E. Since E is in A then I know E is closed under complements and unions. Secondly the integral is with respect to [itex]\mu[/itex], another measure different to [itex]\nu[/itex] (which Im trying to work out if it is a measure).

    So if [itex]\nu[/itex] is going to be a measure then it has to satisfy some properties:

    [tex]1) \quad \nu(\emptyset) = 0[/tex]


    [tex]2) \quad \nu\left(\bigcup_{i=1}^{\infty}E_i\right) = \sum_{i=1}^{\infty}\nu(E_i)[/tex]

    So the first thing I did was see what happens to the empty set. Let me know if this looks right

    [tex]\nu(\emptyset) = \int_{\emptyset}f\mbox{d}\mu[/tex]

    [tex]= \lim_{i\rightarrow\infty}\sum_{i=1}^{\infty}f_i\mu(E)[/tex] where [tex]\lim_{i\rightarrow\infty}f_i = f[/tex]

    [tex]= \lim_{i\rightarrow\infty}\sum_{i=1}^{\infty}f_i\cdot 0[/tex]


    Basically I was looking for a way to get "[itex]\mu(\emptyset)[/itex]" into the picture because I know that the measure of the emptyset is zero - always. So I converted the Lebesgue integral into an equation involving the limit of the sums of a sequence of [itex]f_i[/itex]'s - which involves taking the measure of the set. Im not sure if this is the right way to do it but it is the only way that I could see that I would get zero into the equation.
    Last edited: Sep 17, 2006
  2. jcsd
  3. Sep 17, 2006 #2


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    How have you defined an A-measurable function? Is it that every measurable subset of [itex][0,\infty][/itex] has measurable preimage in X? I'm guesing A is a subset of the power set of X, and consists of all the measurable sets. [itex]\mu[/itex], I'm guessing, is your measure, defined only on A?
    What's with [itex]\mbox{d}\mu[/itex]?
    Huh? It's a Lebesgue integral if you've been studying Lebesgue integrals. Riemann integrals integrate functions over sets as well.
    It doesn't make any sense to say that E is closed under complements and unions. But why is this relevant to evaluating the integral?
    Only if the Ei are disjoint.
    This doesn't make much sense to me at all. What are the fi, what is E, how do the fi (functions?) tend to infinity (a number-like thing), how have you defined the Lebesgue integral?
  4. Sep 17, 2006 #3
    For 2) I guess this would be my first stab in the dark:

    [tex]\nu\left(\bigcup_{i=1}^{\infty}E_i\right) = \int_{E_1\cup E_2\cup\dots}f\mbox{d}\mu[/tex]

    [tex]= \int_{E_1} f\mbox{d}\mu + \int_{E_2}f\mbox{d}\mu + \dots[/tex] (not sure If I can do this step??)

    EDIT: This last step works if f is simple and the E_i's are disjoint. Unfortunately, f needn't be simple. :(

    [tex]=\nu(E_1) + \nu(E_2) \+ \dots[/tex]

    Last edited: Sep 17, 2006
  5. Sep 17, 2006 #4
    Wow, that was quick!

    That's exactly right. Sorry about my apparent vagueness there, I thought that this was the only way of defining an A-measurable function (or at least it is the only way that I have seen). I must be wrong.

    That is exactly how it is written from my source. I guess it means the measure associated with A. That is why this problem concerns me. It had occurred to me that I dont really know how to deal with measures involving integrals (with respect to other measures!)

    Well, I wanted to make sure exactly which kind of sets were being integrated over here - closed sets - since E in A. That is the only reason.

    I guess at this stage I know the Lebesgue integral to be

    [tex]\int_E f\mbox{d}\mu = \int f\chi_E\mbox{d}\mu[/tex]
    Last edited: Sep 17, 2006
  6. Sep 17, 2006 #5
    If my function f is defined as [itex]f\,:\,X\rightarrow[0,\infty][/itex] then does that mean that f is not simple? Since the range of f is infinite?
  7. Sep 17, 2006 #6


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    Another way to define measurability of a function is to consider the undergraph of the function f : X -> [0,oo] in the space X x [0, oo]. If it is a measurable subset of that product space, then f is measurable. When X is Euclidean space, undergraph measurability and preimage measurability coincide, using the standard measures.
    None of that made sense to me.

    Anyways, how does your textbook define the Lebesgue integral, and what theorems do you have?

    There's no reason to assume f is simple.
  8. Sep 18, 2006 #7
    From what I have read, a function is measurable on an interval if there exists a sequence of characteristic functions [itex]\chi[/itex] such that [itex]\chi \rightarrow f[/itex] as n approaches infinity.

    Now what I have is a function, [itex]\nu[/itex] defined on an interval (or a set, or whatever) that is itself an integral. So perhaps I could show that [itex]\nu[/itex] is a measure by showing that there is a sequence of characteristic functions that converge to it. I mean, after all, since the function given is defined in terms of an integral, I should be able to use the functions used to define Upper Riemann sum and show that they converge to it - or something like this?
    Last edited: Sep 18, 2006
  9. Sep 18, 2006 #8


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    You mean a sequence of simple functions?
    Functions are defined on points, and we can talk about functions being measurable. Measures are defined on subsets, and it makes no sense to speak of a measurable/unmeasurable measure.

    Please state precisely how your book defines a measure. It probably says something like:

    m : A -> R+ is a measure iff all the following hold:
    1. m(empty set) = 0
    2. m(disjoint countable union of Ai) = Sum, over i, of m(Ai)

    Then check that v satisfies these properties.
  10. Sep 19, 2006 #9
    - A problem related to "Quantum Physics" (and integration)

    Why is so difficult to find a meassure for "functional integrals " (oo dimensional space) in the form

    [tex] D[\phi]F[\phi] [/tex] ?? and ┬┐Can you use approximate methods to evaluate Lebesgue integrals?..
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