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A measure $\mu$ is a function to $\left[0,\infty\right]$ on the sets belonging to a $\sigma$-algebra.

Then for integrable functions $f$ the integral $\int fd\mu$ comes in, having nice properties like: $$\int f+gd\mu=\int fd\mu+\int gd\mu$$ and $$\int cfd\mu=c\int fd\mu$$ Often I wonder: is there anything that keeps us from defining measure $\mu$ as function on integrable functions (or eventually nonnegative integrable functions) instead of sets?

And in the same line: why not writing $\mu\left(f\right)$ or $\mu f$ instead of $\int fd\mu$? The original value on set $A$ can easily be found back as $\mu\left(1_{A}\right)$.

Are there good reasons not to do this? And if so can you give me some?

drhab
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  • Without the measure as a function defined on a $\sigma$-algebra, how do you determine which functions are integrable? But, one can, on sufficiently nice topological spaces, start with a (continuous) linear form on $C_c(X)$, and then you get a measure from that. – Daniel Fischer Jul 27 '14 at 12:01
  • @DanielFischer Just start with some sort of 'premeasure' on the characteristic functions of measurable sets. Then extend this premeasure to a measure. The premeasure can be used to determine what functions are measurable. In fact the premeasure is the thing what at the present situation is called a measure – drhab Jul 27 '14 at 12:03
  • @DanielFischer I don't plead for throwing things away. It is a matter of renaming and next to that a stimulus to look at integrals instead of measures. They are well-behaved and notations like $\mu f$ are in my view just better than the integral-signs. – drhab Jul 27 '14 at 12:15
  • If you call measures premeasures, and call integrals measures, then what do you call integrals? – Antonio Vargas Jul 27 '14 at 12:34
  • Regarding notation, I have seen $\mu(f)$ instead of $\int f,d\mu$ a couple of times. Nothing wrong with that, although for iterated integrals, or integrals as a function of the upper bound (or upper and lower), the $\int$ notation can be considered to have advantages. Regarding the stimulus to look at integrals instead of measures, apart from some hard-core measure-theorists, doesn't everybody look at measures only as a stepping-stone to looking at integrals? – Daniel Fischer Jul 27 '14 at 12:37
  • @AntonioVargas I admit that not yet everything might be okay yet when it comes to naming. But I am convinced that that will be easy to fix. – drhab Jul 27 '14 at 12:42
  • In fact in some books on Markov Chains (general-space), kernels (of which measures are special instances) are introduced as operators on functions. In particular, in the classical book of Revuz "Markov Chains" there is a precise characterization of such operators that can be represented as kernels. Also I think, these notes may be of interest. – SBF Jul 27 '14 at 12:43
  • @DanielFischer Someone who masters the stuff will off course not have essential profit of a renaming or re-emphasizing. Among students I often see some 'clumseyness' when it comes to integrals that (maybe) would be less if... well I am not sure. – drhab Jul 27 '14 at 12:47
  • @Ilya Thank you. Especially for the link. – drhab Jul 27 '14 at 12:49
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    The http://en.wikipedia.org/wiki/Daniell_integral essentially does what you want. But you apply your approach to arbitrary linear functionals on some vector space of functions. After all, if the functional is given by integration against a measure, then it will be non-negative and have a kind of "monotone convergence" property, see also the link. One can also construct (under suitable hypothesis) a measure that gives back the functional one started with. This is for example done in "Dudley, Real Analysis and Probability". – PhoemueX Jul 27 '14 at 13:04
  • @PhoemueX Thanks. Your comment and especially the link made we aware of my restricted scope. Uptil now I only knew names as Riemann, Stieltjes and Lebesgue in this context. I see now that there is much more. – drhab Jul 27 '14 at 14:58
  • @PhoemueX: care to convert your comment to an answer? – Willie Wong Sep 14 '16 at 05:25

1 Answers1

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First of all, if one wants a "measure defined on a set $\mathcal{F}$ of functions" and if

1) $\mathcal{F}$ only contains functions measurable w.r.t. some $\sigma$-algebra $\mathcal{A}$,

2) $\mathcal{F}$ contains all indicator functions $1_A$ for $A \in \mathcal{A}$,

then it is just more economical to only require a measure to be defined on the sets $A \in \mathcal{A}$ and then use the usual measure-theoretic arguments/theorems to extend the integral.

But of course, there are cases where the two conditions above are not fulfilled, for example one starts from a functional $\mu : C(K) \to \Bbb{R}$, where (e.g.) $C(K)$ are the continuous functions on some compact Hausdorff space $K$. If $\mu(f) \geq 0$ in case of $f \geq 0$, then the Riesz representation theorem shows that there actually is a measure $\nu$ such that $$ \mu(f) = \int f \, d \nu \qquad \forall f \in C(K). $$ The same holds if the space $C(K)$ is replaced by the space $C_c (\Omega)$ of all compactly supported functions on a locally compact Hausdorff space $\Omega$.

There is even a more general construction, the Daniell integral, which is described well in the Wikipedia article and also in Section 4.5 of the (excellent) book "Real Analysis and Probability" by Richard M. Dudley (who calls it the Daniell-Stone integral).

I will follow here the path of Dudley. We start with a so-called vector lattice $F$ of functions $f : X \to \Bbb{R}$ for some base-set $X$. This means that $F$ is a vector space and that if $f,g \in F$, then also $\max\{f,g\} \in F$.

Then, suppose that we are given a linear functional $\mu : F \to \Bbb{R}$ such that

1) If $f \in F$ satisfies $f \geq 0$, then $\mu(f) \geq 0$,

2) If the sequence $(f_n)_n$ in $F$ converges pointwise nonincreasing to $0$, then $\mu(f_n) \to 0$.

Finally, if $F$ is a Stone vector lattice, i.e., if $\min\{f,1\} \in F$ for all $f \in F$, then there is a measure $\nu : \mathcal{A} \to [0,\infty]$, defined on some $\sigma$-algebra $\mathcal{A}$, such that all $f \in F$ are $\mathcal{A}$-measurable and $$ \mu(f) = \int_X f \, d\nu \qquad \forall f \in F. $$

Hence, in most reasonable cases, there is not much difference between the two approaches.

Furthermore, the measure $\nu$ is uniquely determined by the properties above on the smallest $\sigma$-ring for which all $f \in F$ are measurable.

Finally, I have seen the notation $\mu(f)$ some times, in particular (IIRC) in the book "Foundations of modern probability" by Olaf Kallenberg. Thus, it seems to be more or less common in the probabilistic community.

PhoemueX
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