Existence of strictly positive probability measures Let $X$ be a Hausdorff space (or let's even assume it is metrizable). A strictly positive measure on $X$ assigns positive measure to any non-empty open subset of $X$.
What conditions on $X$ ensure the existence of a strictly positive probability (or equivalently $\sigma$-finite) measure? Is there a standard way to construct such a measure if the existence condition is satisfied?
 A: I think a complete characterization might be difficult, so I have collected
some postitive and negative results that should cover most practical cases.


*

*Any compact topological group has a finite Haar measure. If the group
is Hausdorff and infinite this measure is atomless. This is applicable
to e.g. the Cantor space, which can be construed as the product of
$\aleph_0$ copies of the group with two elements.

*If $X$ and $Y$ are Hausdorff spaces and $f: X \to Y$ is continuous,
then for any Borel measure $\mu$ on $X$, a Borel measure $\nu$ on $Y$
can be defined by $\nu(A) = \mu(f^{-1}[A])$ whenever $f^{-1}[A]$ is
$\mu$-measurable. Clearly $\nu(Y) = \mu(X)$, hence if $\mu$ is a
probability measure, so is $\nu$.
a. If $\mu$ is strictly positive and $f[X]$ is dense in $Y$, then $\nu$ is
strictly positive. PhoemueX's answer can be seen as an application of
this where $\mu$ is a geometric measure on $\mathbb{N}$ and $f$ is an
enumeration of a dense subset.
b. If $\mu$ is atomless and $f$ is injective, then $\nu$ is atomless.

*If $\mu$ is an atomless strictly positive Borel probability measure on 
a Hausdorff space $X$,
then for every countable $A \subset X$ we have $\mu(A) = 0$ and the
restriction of $\mu$ to $X\setminus A$ is also an atomless strictly 
positive Borel probability measure. In particular this applies when 
$X = \mathbb{R}, A = \mathbb{Q}$.

*Every Polish space without isolated points has a dense subspace homeomorphic
to $\mathbb{R}\setminus \mathbb{Q}$. (see for example exercise 6.2.A(e) in
Engelking's General topology)  By 2. and 3. any dense embedding of the
irrationals in a space can be used to transfer an arbitrary atomless
strictly positive Borel probability measure from the reals to the 
target space.
The same principle can be used to transfer Lebesgue measure, but the
catch in that case is that the induced measure may not be locally finite.
On the negative side we have:


*If the topology of $Y$ contains an uncountable family of pairwise disjoint
nonempty open sets, there can clearly not be a $\sigma$-finite strictly
positive measure on $Y$. This applies in particular if $Y$ is a
non-separable metrizable space. (see for example Theorem 4.1.15 in
Engelking's General Topology)

*A nonzero measure on a countable set is atomic, since for each point
there is a smallest measurable set that contains it. Consequently, if
a space contains a nonempty countable open set, it does not admit a
strictly positive  atomless measure.



As the part of the exercise referred to in 4. does not come with a reference, I
will give a sketch of how it can be reduced to part (a), which does.
For every separable metric space $X$ and every $i \in \mathbb{Z}^+$ there is a countable family of pairwise disjoint open sets $\{U_{ij}\}_{j=1}^{\infty}$
such that 
$\delta(U_{ij}) \le 2^{-i}$ and $\bigcup_{j=1}^\infty U_{ij}$ is dense in $X$.
By the Baire category theorem, if $X$ is complete we have that
$Y = \bigcap_{i=1}^\infty \bigcup_{j=1}^\infty U_{ij}$ is dense in $X$.
Then, if $X$ has no isolated points, neither has $Y$. It is easy to show
that $\{ U_{ij} \cap Y \mid i, j \in \mathbb{Z}^+ \}$ is a base for the topology
of $Y$ and that all these sets are clopen in $Y$, so $Y$ is zero-dimensional.
By construction $Y$ is $G_\delta$ in $X$, therefore it is completely metrizable.
If we take away a countable dense subset from $Y$ to obtain $Z$, then $Z$
as a subset of $Y$ satisfies the hypothesis of part (a) and is therefore
homeomorphic to the irrationals.
An direct construction of a suitable measure might go along similar lines.
If $\{U_{ij}\}$ is constructed in such a way that for every $j$ and every
$m < n$ there is a $k$ such that $U_{nj} \subset U_{mk}$, it is possible
to assign a positive number to each $U_{ij}$ in such a way that countable
additivity is satisfied. Such an assignment can be extended to a strictly
positive measure, and by requiring that each point is contained in sets
of arbitrarily small measure, it can be guaranteed that the resulting 
measure is atomless.
A: If there is a countable dense subset $\{x_n \mid n \in \Bbb{N}\}$, we can define $\mu =\sum_n 2^{-n} \delta_{x_n}$, where $\delta_{x_n}$ is the dirac measure (point mass) at $x_n$.
As every open set contains some $x_n$, this does what you want.
