# Equality of topologies on the space of positive Radon measures

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.

• Is this true for compact $X$? – Sergei Akbarov Feb 14 '20 at 20:17
• @SergeiAkbarov I added the proof for compact $X$. – yada Feb 15 '20 at 7:51
• 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 $f(x)\mu(dx)$. – Sergei Akbarov Feb 15 '20 at 8:46
• For measures with density I see people often writing $f \mu$. – yada Feb 15 '20 at 11:01
• If they distinguish $f\mu$ and $\mu f$, this also looks confusing. – Sergei Akbarov Feb 15 '20 at 13:40

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 $$\mu = 0$$).