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Let $X$ be a locally compact, Hausdorff, topological space and denote by $\mathcal B_+(X)$ the collection of all Borel-measurable functions from $X$ to $[0,+\infty]$ (extended positive reals).

Suppose that we are given a map $$ I:\mathcal B_+(X) \to [0,+\infty] $$ which is positively homogeneous and satisfies $$ I\big(\sum_{n=1}^\infty f_n\big) = \sum_{n=1}^\infty I(f_n), $$ for every sequence $\{f_n\}_n$ in $\mathcal B_+(X)$.

Observe that this is essentially the Daniell integral.

Suppose further that $\mu$ is a given regular Borel measure on $X$ such that $$ I(f) = \int_X f\,d\mu, $$ for every non-negative $f$ in $C_c(X)$, the space of all compactly supported continuous functions on $X$. Does it follow that this identity also holds for every $f$ in $\mathcal B_+(X)$?

Ok to assume that $X$ is metrizable and $\sigma$-compact, if necessary.

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  • $\begingroup$ It is easy to prove that the identity above holds for every measurable function which is the pointwise limit of an increasing sequence $\{f_n\}_n$ in $C_c(X)$ but, unfortunately, not every $f$ in $\mathcal B_+(X)$ is of this form. $\endgroup$
    – Ruy
    Commented May 11, 2018 at 0:12
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    $\begingroup$ I think this should be a typical exercise with the monotone class theorem or some variant. $\endgroup$
    – Nate Eldredge
    Commented May 11, 2018 at 3:39
  • $\begingroup$ By the way, you do need the assumption of metrizability (or something similar): otherwise, let $X = \omega_1$ and let $I$ be the integral with respect to the Dieudonné measure (a set has measure 1 if it contains a club and 0 otherwise). The integral of every continuous compactly supported function is zero, so $\mu=0$ satisfies the hypotheses, but not the conclusion. You can even have a compact example by looking at $\omega_1 + 1$; here $I$ could be integration against the Dieudonné measure on $\omega_1$ and $\mu$ a point mass at the endpoint. $\endgroup$
    – Nate Eldredge
    Commented May 12, 2018 at 12:35
  • $\begingroup$ @Nate, indeed this is a nice example. I always feel a bit lazy when assuming so many countability hypotheses but your example does show that life is impossible without them. I am currently writing up a paper in which these ideas are relevant and I am settling for $\sigma$-compact spaces and, most importantly, Baire measures. I think it is a big mistake to give so much emphasis to Borel measures when the main functions you want to integrate are compactly supported continuous functions! $\endgroup$
    – Ruy
    Commented May 12, 2018 at 22:46

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Here is an answer for $X$ metrizable and $\sigma$-compact. I assume $I$ is finite on compactly supported continuous functions.

Recall that a locally compact metrizable $\sigma$-compact space is Polish (separable and completely metrizable). Moreover, there is an increasing sequence of compact sets $\langle K_n\rangle$ such that $K_n$ is a subset of the interior of $K_{n+1}$ for all $n$ and such that $X=\bigcup_n K_n$. Fix some metric $d$ that completely metrizes $X$.

Clearly, it suffices to show $I(1_A)=\mu(A)$ for every measurable set $A$. By the regularity of $\mu$, one only has to prove this for $A$ compact and, even nicer, for sets of the form $A\cap K_n$ with $A$ compact. Now there is going to be some $\epsilon$ such that $(A\cap K_n)_\epsilon=\{y\in X\mid d(y, A\cap K_n)<\epsilon\}$ is a subset of the interior of $K_{n+1}$. Define $f_n:X\to\mathbb{R}$ by $f_m(x)=\max\{1-md(x,A\cap K_n),0\}$. This is a continuous function and for $m$ larger than $1/\epsilon$, it vanishes outside the compact set $K_{n+1}$. Moreover $\langle f_m\rangle$ decreases pointwise to $1_{A\cap K_n}$. Since $I$ satisfies the dominated convergence theorem, the conclusion follows.

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  • $\begingroup$ Thanks for the nice answer. I was missing the dominated convergence Thm! $\endgroup$
    – Ruy
    Commented May 11, 2018 at 17:15
  • $\begingroup$ I guess your proof easily generalizes w/out the $\sigma$ compactness. Given any compact $K$, the local compactness of $X$ guarantees that $K$ is contained in the interior of another compact set. $\endgroup$
    – Ruy
    Commented May 11, 2018 at 17:18
  • $\begingroup$ @Ruy That's right. $\endgroup$
    – Michael Greinecker
    Commented May 11, 2018 at 17:28
  • $\begingroup$ One more point: my way to see that $I$ satisfies the dominated convergence Theorem is a bit roundabout, going through the measure defined by $I$. Can you prove it in a more direct way? $\endgroup$
    – Ruy
    Commented May 11, 2018 at 17:44
  • $\begingroup$ @Ruy (Increasing) monotone convergence is fairly straightforward for $I$ and you get downward monotone convergence for $\langle f_n\rangle$. The domination is only needed so you can transform this into increasing monotone convergence by subtracting the $f_n$ from an integrable function and then switching things around in the end. Still a bit roundabout though. $\endgroup$
    – Michael Greinecker
    Commented May 11, 2018 at 17:59

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