Scaling Property in DFT
$\sum\limits_{m=0}^{M-1} \sum\limits_{n=0}^{N-1} f(am,bn)e^{-j \cdot 2\pi(\frac {k}{M} \cdot m + \frac {l}{N} \cdot n)} $  (let $m'=am, n'=bn$)

$=\sum\limits_{m'=0}^{(M-1)a} \sum\limits_{n'=0}^{(N-1)b} f(m',n')e^{-j \cdot 2\pi(\frac {k/a}{M} \cdot m' + \frac {l/b}{N} \cdot n')}$  

($m'$, $n'$ with the incremental interval of $a$ and $b$ respectively)

Note that 
$\sum\limits_{m'=0}^{M-1} \sum\limits_{n'=0}^{N-1} f(m',n')e^{-j \cdot 2\pi(\frac {k/a}{M} \cdot m' + \frac {l/b}{N} \cdot n')}$
$=F(k/a,l/b)$

Here $m'$, $n'$ with the incremental interval of $1$, and the points $m$ and $n$ are evenly distributed on the circle, according to the symmetry, you have 

$\sum\limits_{m'=0}^{(M-1)a} \sum\limits_{n'=0}^{(N-1)b} f(m',n')e^{-j \cdot 2\pi(\frac {k/a}{M} \cdot m' + \frac {l/b}{N} \cdot n')} = \frac {1}{|ab|}F(k/a,l/b)$