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Every $\sigma$-smooth measure is $\tau$-smooth. This is what we need. As noted, if there is a (real-valued) measurable cardinal, then this may fail for a metric space. A space is called "measure-compact" iff every $\sigma$-smooth measure is $\tau$-smooth.

The reference for all of this (up to 1960) is: Gillman & Jerison, RINGS OF CONTINUOUS FUNCTIONS. In a completely regular space we would use "zero sets" (a set where some continuous real-valued function vanishes). But in a metric space these are the same as the closed sets. A (finite, Borel) measure $\mu$ on a metric space is $\sigma$-smooth iff it is coutably additive, but this means if $A_n$ is a decreasing sequence of closed sets, then $\mu(A_n)$ converges to $\mu(\bigcap_n A_n)$. A stronger condition on $\mu$ is $\tau$-smooth: if $A_t$ is a decreasing net of closed sets, then $\mu(A_t)$ converges to $\mu(\bigcap_t A_t)$. The "support" of a probability measure $\mu$ is the intersection of all closed sets of measure $1$. And (assuming $\mu$ is $\tau$-smooth) this intersection again has measure $1$.

As I recall, a metric space is measure-compact if and only if there is no discrete subset with real-valued measurable cardinal. So, in particular, if there are no real-valued measurable cardinals, then the answer to the question in the title is YES. Joel has provided the converse. Thus this question is presumably independent of ZFC.

The term "measure-compact" is due to Moran, 1965. By analogy with "real-compact" which may be characterized in the same way using only $\{0,1\}$-valued measures.

Every $\sigma$-smooth measure is $\tau$-smooth. This is what we need. As noted, if there is a (real-valued) measurable cardinal, then this may fail for a metric space. A space is called "measure-compact" iff every $\sigma$-smooth measure is $\tau$-smooth.

The reference for all of this (up to 1965) is: V. S. Varadarajan, "Measures on Topological Spaces". In a completely regular space we would use "zero sets" (a set where some continuous real-valued function vanishes). But in a metric space these are the same as the closed sets. A (finite, Borel) measure $\mu$ on a metric space is $\sigma$-smooth iff it is coutably additive, but this means if $A_n$ is a decreasing sequence of closed sets, then $\mu(A_n)$ converges to $\mu(\bigcap_n A_n)$. A stronger condition on $\mu$ is $\tau$-smooth: if $A_t$ is a decreasing net of closed sets, then $\mu(A_t)$ converges to $\mu(\bigcap_t A_t)$. The "support" of a probability measure $\mu$ is the intersection of all closed sets of measure $1$. And (assuming $\mu$ is $\tau$-smooth) this intersection again has measure $1$.

As I recall, a metric space is measure-compact if and only if there is no discrete subset with real-valued measurable cardinal. So, in particular, if there are no real-valued measurable cardinals, then the answer to the question in the title is YES. Joel has provided the converse. Thus this question is presumably independent of ZFC.

The term "measure-compact" is due to Moran, 1965. By analogy with "real-compact" which may be characterized in the same way using only $\{0,1\}$-valued measures.

Every $\sigma$-smooth measure is $\tau$-smooth. This is what we need. As noted, if there is a (real-valued) measurable cardinal, then this may fail for a metric space. A space is called "measure-compact" iff every $\sigma$-smooth measure is $\tau$-smooth.

The reference for all of this (up to 1960) is: Gillman & Jerison, RINGS OF CONTINUOUS FUNCTIONS. In a completely regular space we would use "zero sets" (a set where some continuous real-valued function vanishes). But in a metric space these are the same as the closed sets. A (finite, Borel) measure $\mu$ on a metric space is $\sigma$-smooth iff it is coutably additive, but this means if $A_n$ is a decreasing sequence of closed sets, then $\mu(A_n)$ converges to $\mu(\bigcap_n A_n)$. A stronger condition on $\mu$ is $\tau$-smooth: if $A_t$ is a decreasing net of closed sets, then $\mu(A_t)$ converges to $\mu(\bigcap_t A_t)$. The "support" of a probability measure $\mu$ is the intersection of all closed sets of measure $1$. And (assuming $\mu$ is $\tau$-smooth) this intersection again has measure $1$.

As I recall, a metric space is measure-compact if and only if there is no discrete subset with real-valued measurable cardinal. So, in particular, if there are no real-valued measurable cardinals, then the answer to the question in the title is YES. Joel has provided the converse. Thus this question is presumably independent of ZFC.

The term "measure-compact" dates from later, so it is not found in Gillman & Jerison.

Every $\sigma$-smooth measure is $\tau$-smooth. This is what we need. As noted, if there is a (real-valued) measurable cardinal, then this may fail for a metric space. A space is called "measure-compact" iff every $\sigma$-smooth measure is $\tau$-smooth.

The reference for all of this (up to 1960) is: Gillman & Jerison, RINGS OF CONTINUOUS FUNCTIONS. In a completely regular space we would use "zero sets" (a set where some continuous real-valued function vanishes). But in a metric space these are the same as the closed sets. A (finite, Borel) measure $\mu$ on a metric space is $\sigma$-smooth iff it is coutably additive, but this means if $A_n$ is a decreasing sequence of closed sets, then $\mu(A_n)$ converges to $\mu(\bigcap_n A_n)$. A stronger condition on $\mu$ is $\tau$-smooth: if $A_t$ is a decreasing net of closed sets, then $\mu(A_t)$ converges to $\mu(\bigcap_t A_t)$. The "support" of a probability measure $\mu$ is the intersection of all closed sets of measure $1$. And (assuming $\mu$ is $\tau$-smooth) this intersection again has measure $1$.

As I recall, a metric space is measure-compact if and only if there is no discrete subset with real-valued measurable cardinal. So, in particular, if there are no real-valued measurable cardinals, then the answer to the question in the title is YES. Joel has provided the converse. Thus this question is presumably independent of ZFC.

The term "measure-compact" is due to Moran, 1965. By analogy with "real-compact" which may be characterized in the same way using only $\{0,1\}$-valued measures.

Every $\sigma$-smooth measure is $\tau$-smooth. This is what we need. As noted, if there is a (real-valued) measurable cardinal, then this may fail for a metric space. A space is called "measure-compact" iff every $\sigma$-smooth measure is $\tau$-smooth.

The reference for all of this (up to 1960) is: Gillman & Jerison, RINGS OF CONTINUOUS FUNCTIONS. In a completely regular space we would use zero sets" (a set where some continuous real-valued function vanishes). Butin a metric space these are the same as the closed sets. A (finite, Borel) measure $\mu$ on a metric space is $\sigma$-smooth iff it is coutably additive, but this means if $A_n$ is a decreasing sequence of closed sets, then $\mu(A_n)$ converges to $\mu(\bigcap_n A_n)$. A stronger condition on $\mu$ is $\tau$-smooth: if $A_t$ is a decreasing net of closed sets, then $\mu(A_t)$ converges to $\mu(\bigcap_t A_t)$. The "support" of a probability measure $\mu$ is the intersection of all closed sets of measure $1$. And (assuming $\mu$ is $\tau$-smooth) this intersection again has measure $1$.

As I recall, a metric space is measure-compact if and only if there is no discrete subset with real-valued measurable cardinal. So, in particular, if there are no real-valued measurable cardinals, then the answer to the question in the title is YES. Joel has provided the converse. Thus this question is presumably independent of ZFC.

Every $\sigma$-smooth measure is $\tau$-smooth. This is what we need. As noted, if there is a (real-valued) measurable cardinal, then this may fail for a metric space. A space is called "measure-compact" iff every $\sigma$-smooth measure is $\tau$-smooth.

The reference for all of this (up to 1960) is: Gillman & Jerison, RINGS OF CONTINUOUS FUNCTIONS. In a completely regular space we would use "zero sets" (a set where some continuous real-valued function vanishes). But in a metric space these are the same as the closed sets. A (finite, Borel) measure $\mu$ on a metric space is $\sigma$-smooth iff it is coutably additive, but this means if $A_n$ is a decreasing sequence of closed sets, then $\mu(A_n)$ converges to $\mu(\bigcap_n A_n)$. A stronger condition on $\mu$ is $\tau$-smooth: if $A_t$ is a decreasing net of closed sets, then $\mu(A_t)$ converges to $\mu(\bigcap_t A_t)$. The "support" of a probability measure $\mu$ is the intersection of all closed sets of measure $1$. And (assuming $\mu$ is $\tau$-smooth) this intersection again has measure $1$.

As I recall, a metric space is measure-compact if and only if there is no discrete subset with real-valued measurable cardinal. So, in particular, if there are no real-valued measurable cardinals, then the answer to the question in the title is YES. Joel has provided the converse. Thus this question is presumably independent of ZFC.

The term "measure-compact" dates from later, so it is not found in Gillman & Jerison.