Inequality involving counting measure

Question

Can someone explain to me how the following inequality is true ??

for each $k \in \mathbb N \cup $ {$0$} and any $r \in (0,\infty)$,

#$_{\mathbb N_{0}}(${$ n \in \mathbb N_{0}: k \leq nr \lt k+1$}$) \quad \leq $ $\quad$ #$_{\mathbb N_{0}}(${$ n \in \mathbb N_{0}: nr \lt 1$}$)$

-where # denotes the counting measure!!

Well, let me clarify, I was reading some stuffs on Gamma Function and in the proof of one of the inequality, the argument used was:

$\sum_{k=0}^{\infty} \frac{x^{k}}{\Gamma(k)}$ #$_{\mathbb N_{0}}(${$ n \in \mathbb N_{0}: k \leq nr \lt k+1$}) $\leq$ #$_{\mathbb N_{0}}(${$ n \in \mathbb N_{0}: nr \lt 1$}$) \sum_{k=0}^{\infty} \frac{x^{k}}{\Gamma(k)}$

Well, I get the case of equality but in general when $r \in (0, \infty)$ how does the inequality is coming ??

Thanks in advance!!

Answer 1

For fixed $k \in \Bbb N_0$ and $r \in (0, \infty)$ denote $$ A = \{ n \in \mathbb N_{0}: k \leq nr \lt k+1 \} \\ B = \{ n \in \mathbb N_{0}: nr \lt 1 \} \\ m = \min A \, . $$ For each $n \in A$, $$ 0 \le (n-m)r = nr - mr < (k+1) - k = 1 $$ which means that $f(n) = n - m$ is an injective mapping from $A$ to $B$, and therefore $B$ has at least as many elements as $A$.

Intuitively, we shift all the numbers $nr$ in the interval $[k, k+1)$ by a constant amount to the left, so that the smallest one becomes zero. The shifted numbers are then all in the interval $[0, 1)$.