Why does the Continuum Hypothesis make an ideal measure on $\mathbb R$ impossible?

This is a consequence of Ulam's theorem: If a finite countable additive measure $\mu$ is defined on all subsets of a set of cardinality $\aleph_1$ (the first uncountable cardinal) and vanishes on all singletons, then it is identically zero.

This is the version of the theorem from the book of Bogachev (1.12.40), the original result is a bit stronger. Anyways, under CH, $\mathbb{R}$ has cardinality $\aleph_1$, so the theorem applies to $\mathbb{R}$. I'm pretty sure this is what Royden refers to.

I refer you to the MathOverflow question "Does pointwise convergence imply uniform convergence on a large subset?" for proofs and references for the following:

Assuming the Continuum Hypothesis, there exists a sequence of real-valued functions on $\mathbb R$ that converges pointwise to a function on $\mathbb R$ but does not converge uniformly on any uncountable set.

Suppose there were a function $m:\mathcal{P}(\mathbb R)\to[0,\infty]$ satisfying the three named conditions. Then $m$ is a measure on $\mathbb R$ for which every subset of $\mathbb R$ is measurable. Let $(f_n)$ be a sequence of real-valued functions on $[0,1]$ that converges pointwise to some function. By Egorov's theorem, there is a set $E\subseteq [0,1]$ with $m(E)>0$ such that $(f_n)$ converges uniformly on $E$. The fact that $m(E)>0$ implies that $E$ is uncountable. Since $(f_n)$ was arbitrary, this implies by the above cited result that the Continuum Hypothesis does not hold.

In light of Michael's answer, I want to mention that the hypothesis 2, that $m(I)=l(I)$ for each interval $I$, could be replaced by the following two properties:

(a) For all $x\in\mathbb R$, $m(\{x\})=0$.

(b) There exists $X\subseteq\mathbb R$ such that $0<m(X)<\infty$.

Then the same argument as above would apply with such an $X$ in place of $[0,1]$.