The "Hard measure problem on $\mathbb{R}$" is to find a nonnegative set theoretic function $\rho$ whose domain are subsets of $\mathbb{R}$, such that:
- $\rho(E)$ is defined for all $E\subseteq\mathbb{R}$;
- $\rho(I) = \mathrm{length}(I)$ for all intervals $I\subseteq\mathbb{R}$;
- $\rho$ is $\sigma$-additive;
- $\rho$ is invariant under isometries; i.e., if $j\colon\mathbb{R}\to\mathbb{R}$ is an isometry, then $\rho(j(E)) = \rho(E)$ for all $E\subseteq $\mathbb{R}.
If we assume the Axiom of Choice, then the hard measure problem on \mathbb{R} cannot be solved, as shown by the construction of Vitali sets. On the other hand, Solovay proved that in there is a model of Set Theory without the Axiom of Choice in which the hard problem can be solved.
The "Easy measure problem on \mathbb{R}$" requires a measure that satisfies 1, 2, and 4 above, and replaced 3 with simple finite additivity. Banach proved that the problem can be solved (non-uniquely) in $\mathbb{R}$, and in $\mathbb{R}^2$ (where 2 replaces interval and length with rectangle and area), and Hausdorff proved that the easy measure problem cannot be solved in $\mathbb{R}^n$ with $n\geq 3.
The does not directly address your question, because a measure that satisfies 2 above cannot have \rho(\mathbb{R})=1. H
A probability measure is just a function that takes values on [0,1]$, is $\sigma$-additive, and satisfies $\rho(\emptyset)=0$ and $\rho(\mathbb{R})=1$. There are plenty of probability measures on all of $\mathbb{R}$ that are defined on every subset: you can define atomic measures in which there are countably many atoms and their measures add up to $1$, for instance. However, usually you want some sort of "uniformity" to the measure (e.g., there is no uniform probability measure on $\mathbb{N}$, or on any countable set). So one needs to specify exactly what conditions you want to place on the measure in order to determine whether or not one can construct such a probability measure.