Cardinality of the set of increasing real functions Could you show me how to "calculate" the cardinality of the set of increasing (not necessarily strictly) functions $\ f: \mathbb{R} \rightarrow \mathbb{R}$ ?
 A: There are only $|\Bbb R|=2^\omega$ such functions.
Let $\varphi:\Bbb Q\to\Bbb R$ be non-decreasing. There are only $|\Bbb R|^{|\Bbb Q|}=\left(2^\omega\right)^\omega=2^\omega$ functions from $\Bbb Q$ to $\Bbb R$, and it’s easy to see that there are at least $|\Bbb R|=2^\omega$that are non-decreasing, so there are $2^\omega$ such functions $\varphi$. If $f:\Bbb R\to\Bbb R$ is non-decreasing, then $f\upharpoonright\Bbb Q$ is one of these $2^\omega$ functions, so we’d like to know how many non-decreasing functions from $\Bbb R$ to $\Bbb R$ restrict to a given non-decreasing $\varphi:\Bbb Q\to\Bbb R$.
Let $\varphi:\Bbb Q\to\Bbb R$ be non-decreasing, and suppose that $f:\Bbb R\to\Bbb R$ is non-decreasing and restricts to $\varphi$ on $\Bbb Q$. For each irrational $x$ let 
$$\varphi^-(x)=\sup_{q\in\Bbb Q\cap(\leftarrow,x)}\varphi(q)$$
and
$$\varphi^+(x)=\inf_{q\in\Bbb Q\cap(x,\to)}\varphi(q)\;;$$
then $\varphi^-(x)\le f(x)\le \varphi^+(x)$. If $\varphi^-(x)=\varphi^+(x)$, then there is only one way to define $f(x)$. Otherwise, there are $\left|\big[\varphi^-(x),\varphi^+(x)\big]\right|=2^\omega$ choices for $f(x)$. 
Let $$C=\left\{x\in\Bbb R\setminus\Bbb Q:\big(\varphi^-(x),\varphi^+(x)\big)\ne\varnothing\right\}\;;$$ the intervals $\big(\varphi^-(x),\varphi^+(x)\big)$ for $x\in C$ are pairwise disjoint, so there are at most countably many of them. Thus, $f(x)$ is completely determined by $\varphi$ except on the countable set $C$, and for each $x\in C$ there are $2^\omega$ possible values for $f(x)$, so there are at most $\left(2^\omega\right)^\omega=2^\omega$ possible non-decreasing functions $f:\Bbb R\to\Bbb R$ such that $f\upharpoonright\Bbb Q=\varphi$.
Putting the pieces together, we see that there are at most $2^\omega\cdot2^\omega=2^\omega$ non-decreasing functions $f:\Bbb R\to\Bbb R$, and the constant functions already show that there are at least that many.
A: Any nondecreasing function is Borel measureable, but there are only $\mathbf{R}$ many Borel measureable functions.
