Let $X$ be a locally compact Hausdorff space, $C_0(X)$ the Banach space of continuous functions vanishing at infinity, $M(X) := C_0(X)'$ the space of Radon measures and $M^+(X) \subseteq M(X)$ the positive finite Radon measures. On $M(X)$, denote by $w^*$ the weak$^*$ topology (relative to $C_0(X)$) and by $\tau$ the topology of uniform convergence on norm compact sets of $C_0(X)$, so that $w^* \subseteq \tau$. It is known that $\tau$ coincides with the topology of uniform convergence on norm null sequences (by a theorem of Grothendieck, every norm compact set in a Banach space is contained in the absolutely convex closure of a norm null sequence).

Is it true that on $M^+(X)$ it holds $w^* = \tau$?

So, we have to show that for nets $\mu_\alpha, \mu \in M^+(X)$ with $\mu_\alpha f \to \mu f$ for each $f \in C_0(X)$ it also holds $\sup_n |(\mu_\alpha - \mu) f_n| \to 0$ for each sequence $f_n \in C_0(X)$ with $f_n \geq 0$ and $\lVert f_n \rVert \to 0$.

I have read somewhere that this is true for a compact space $X$. So, it may be also true for a locally compact Hausdorff space. But since the above condition involves sequences, I think, one has to restrict to $\sigma$-compact or paracompact spaces $X$.

Edit: Here is the proof for compact $X$:

Let $\mu_\alpha \to \mu$ for $w^*$ in $M^+(X)$. Since $X$ is compact, $1_X \in C_0(X) = C(X)$. From $\mu_\alpha 1_X \to \mu 1_X$, there is $\alpha_0$ such that $0 \leq \mu_\alpha 1_X \leq \mu 1_X + 1 =: c$ for all $\alpha \geq \alpha_0$. Then for any $f \in C(X)$: $|\mu_\alpha f| \leq \lVert \mu_\alpha \rVert \cdot \lVert f \rVert = \mu_\alpha 1_X \cdot \lVert f \rVert \leq c \lVert f \rVert$ for all $\alpha \geq \alpha_0$. Therefore, $\{ \mu_\alpha \mid \alpha \geq \alpha_0 \}$ is $w^*$-bounded. By Banach-Alaouglu, this set is $w^*$-relatively compact and since $\tau$ and $w^*$ coincide on $w^*$-compact sets (because $C(X)$ is complete) it follows that $\mu_\alpha \to \mu$ for $\tau$.

For non-compact $X$, I think, $1_X$ should be replaced by some strictly positive function in $C_0(X)$, and these do exist, if $X$ is paracompact - have to think about it.

  • $\begingroup$ Is this true for compact $X$? $\endgroup$ Feb 14, 2020 at 20:17
  • $\begingroup$ @SergeiAkbarov I added the proof for compact $X$. $\endgroup$
    – yada
    Feb 15, 2020 at 7:51
  • $\begingroup$ yadaddy, it seems to me it will better to write $\mu(f)$ instead of $\mu f$, because the latter can be confused with the measure $f(x)\mu(dx)$. $\endgroup$ Feb 15, 2020 at 8:46
  • $\begingroup$ For measures with density I see people often writing $f \mu$. $\endgroup$
    – yada
    Feb 15, 2020 at 11:01
  • $\begingroup$ If they distinguish $f\mu$ and $\mu f$, this also looks confusing. $\endgroup$ Feb 15, 2020 at 13:40

1 Answer 1


For the case $\mu = 0$ one can proceed as follows.

For a given sequence $f_n \in C_0(X)$, $f_n \geq 0$ with $\lVert f_n \rVert \to 0$ construct a function $g \in C_0(X)$ such that $f_n \leq g$ for all $n$. Then $|\mu_\alpha f_n| = \mu_\alpha f_n \leq \mu_\alpha g$ for each $n$ since $\mu_\alpha \geq 0$ and $f_n \geq 0$. It follows that $\sup_n |\mu_\alpha f_n| \leq \mu_\alpha g \to 0$.

Construction of $g$: From $\lVert f_n \rVert = \sup_{x \in X} |f_n(x)| \to 0$ we can iteratively construct a sequence of indices $0 \leq n_1 < n_2 < n_3 < \dots$ such that $\lVert f_n \rVert \leq \frac{1}{k}$ for all $n \geq n_k$.

(0) For the finite initial part $f_0, \dots, f_{n_1-1} \in C_0(X)$ there is a compact $K_0 \subseteq X$ such that $f_0, \dots, f_{n_1-1} \leq 1$ on $X \setminus K_0$ and $\leq M$ on $K_0$ for some $M \geq 1$. Define $g_0(x) := M$ for all $x \in X$. Then $f_n \leq g_0$ for all $n \in \mathbb{N}$.

(1) For $n \geq n_1$ we know that $f_n \leq 1$ on $X$. Take any relatively compact open $U_0 \supseteq K_0$. Define $g_1 : X \to \mathbb{R}$ as follows. For $x \in K_0$ set $g_1(x) := g_0(x)$. For $x \in X \setminus U_0$ set $g_1(x) := 1$. Extend the so-defined function $g_1$ on $K_0 \cup (X \setminus U_0)$ to a continuous function $g_1$ defined on $X$ satisfying $1 \leq g_1 \leq g_0$ as follows: there is a continuous function $\psi_1 : X \to \mathbb{R}$ such that $\psi_1 = 1$ on $K_0$, $\psi_1 = 0$ on $X \setminus U_0$ and $0 \leq \psi_1 \leq 1$. Then $g_1(x) := g_0(x) \cdot \psi_1(x) + 1 \cdot (1-\psi_1(x))$ defined for $x \in X$ is the desired continuous extension. It holds $1 \leq g_1 \leq g_0$ on $X$, $g_1 = g_0$ on $K_0$ and $g_1 = 1$ on $X \setminus U_0$. Observe that $f_n \leq g_1$ on $X$ for all $n \in \mathbb{N}$.

(2) For $n \geq n_2$ we know that $f_n \leq \frac{1}{2}$ on $X$. For the finite collection $f_0, \dots, f_{n_2-1} \in C_0(X)$ there is a compact $K_1 \subseteq X$ such that $f_0, \dots, f_{n_2-1} \leq \frac{1}{2}$ on $X \setminus K_1$. We can assume that $U_0 \subseteq K_1$ by potentially enlarging $K_1$. Take any relatively compact open neighborhood $U_2$ of $K_1$. Define $g_2 : X \to \mathbb{R}$ as follows. For $x \in K_1$ set $g_2(x) := g_1(x)$. For $x \in X \setminus U_1$ set $g_2(x) := \frac{1}{2}$. Extend the so-defined function $g_2$ on $K_1 \cup (X \setminus U_1)$ to a continuous function $g_2$ defined on $X$ satisfying $\frac{1}{2} \leq g_2 \leq g_1$ as in step (1). Observe that $f_n \leq g_2$ on $X$ for all $n \in \mathbb{N}$. In fact, from $f_n \leq g_1$ on $X$ for all $n \in \mathbb{N}$ we get $f_0, \dots, f_{n_2-1} \leq g_1 = g_2$ on $K_1$, so that with $f_0, \dots, f_{n_2-1} \leq \frac{1}{2} \leq g_2$ on $X \setminus K_1$ we get $f_0, \dots, f_{n_2-1} \leq g_2$ on $X$. For $n \geq n_2$ we already know that $f_n \leq \frac{1}{2} \leq g_2$ on $X$.

We can now proceed iteratively. This yields a sequence $g_k \in C_b(X)$ satisfying $0 \leq g_k \leq M$ on $X$. Since $g_k$ is pointwise decreasing it follows that $g_k(x)$ converges for any $x \in X$. Define $g(x) := \lim_{k \to \infty} g_k(x)$ for any $x \in X$. From $f_n \leq g_k$ on $X$ for all $n$ and all $k$ it follows that $f_n \leq g$ for all $n$. Finally, to see that $g \in C_0(X)$ let $\varepsilon > 0$ and take any $k > \frac{1}{\varepsilon}$. By construction, we have $g_k = \frac{1}{k}$ on $X \setminus U_{k-1}$. Then from $K_k \supseteq U_{k-1}$ it follows that $g \leq g_k = \frac{1}{k} < \varepsilon$ on $X \setminus K_k$.

EDIT: It would be interesting to know whether a similar proof applies for a general $\mu \geq 0$ - I expect that we then need also to approximate the integrals $\mu f$ in a suitable way (this is obviously not necessary for the case $\mu = 0$).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .