When we define double integral for function $f$, we usually start with it's definition on a rectangle $R=[a,b]\times[c,d]$ which can be split into bunch of smaller rectangles $[a,b]=[x_0,x_1]\cup[x_1,x_2]\cup...\cup[x_{n-1},x_n]$ and $[c,d]=[y_0,y_1]\cup[y_1,y_2]\cup...\cup[y_{n-1},y_n]$ then forming a sum
where $\Delta x_i=x_i-x_{i-1}$ and $\Delta y_j=y_j-y_{j-1}$ and $(\epsilon_i, \gamma_j)$ is point inside rectangle $R_{ij}$
Now, if we choose the rectangle with greatest diameter (let that diameter be $d_{ij}$ ) and if we what to find limit of the given sum when $d_{ij} \rightarrow 0 $, then if that sum has a finite limit regardless of how we split the intervals $[a,b]$ and $[c,d]$ and which points inside $R_{ij}$ rectangles we choose, then, that limit of the sum is called DOUBLE INTEGRAL of $f$ on $R$.
However, i am supposed to understand the definition of double integral in $\epsilon - \delta$ terms.
Defining double integral this way makes sense since this is obviously a limit and we know that limits can be definied in $\epsilon - \delta$ terms.
There it goes:
Number $I$ is called double integral of function $f$ on rectangle $R$ if for every $\epsilon > 0$ there is $\delta >0$ such that if we split intervals $[a,b]$ and $[c,d]$ and we get $d_{ij}<\delta$, then, regardless of points chosen following inequality is true:
$$| \sum_{i=1}^m \sum_{j=1}^n f(\epsilon_i, \gamma_j)\Delta x_i \Delta y_j - I | < \epsilon$$
Now, this should't be difficult to understand, but I struggle with it very much, since i cannot understand why is it required that we have to split given intervals in such way that $d_{ij}<\delta$, why this is not working if we split the intervals in a such way that $d_{ij}>\delta$ (AGAIN: $d_{ij}$ is a diameter of largest rectangle we got when we split the two intervals given). Any help is appreciated!
It's not clear what you mean by “not working” here. Do you mean, why is it necessary that $d<\delta$? Otherwise, the condition is not equivalent to the definition.
I mean, it's possible that for some $\epsilon$, there exists a partition with $d > \delta$ and the difference between the Riemann sum and $I$ is smaller than $\epsilon$. But if you are looking for functions which satisfy that condition for all partitions with $d>\delta$, you're going to eliminate a lot of integrable functions.
Compare it to the $\epsilon$-$\delta$ definition of limit. We say $\lim_{x\to a}f(x) = L$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that $0 < | x- a | < \delta \implies |f(x) - L| < \epsilon$. Why do we rule out points $x$ with $|x - a| > \delta$? Without it, the statement that $\lim_{x\to a} x = a$ wouldn't be true.