What happens if you integrate over a space filling curve?

Question

So I was thinking about space-filling curves. For simplicity, let us say that we wish to fill the unit square with a curve. From my understanding, one way to think of space-filling curves would be to be a sequence of curves $\{\gamma_i\}_{i=1}^\infty$ where $\gamma_i:[0,1]\rightarrow[0,1]^2$ is a curve such that for every $x\in [0,1]^2$ there is some $N$ such that for $n>N$ there is some $t\in[0,1]$ such that $\gamma_n(t)=x$. Now let us say that I have a function $f:[0,1]^2\rightarrow \mathbb{R}$. Then I had thought that it would make sense that the integral $$ \int_{[0,1]^2}fdA=\lim_{n\rightarrow\infty}\int_{\gamma_n}fdt $$ That is it seemed intuitive to me that integrating over the entire region should be the same as the limit of integrating over the space-filling curve. However, the more I thought about it, the more I began to question myself as on the left-hand side we are integrating the area while on the right-hand side we are integrating over a curve. Thus, if we assume that $f$ is continuous and positive, then it attains a minimum and maximum value on $[0,1]^2$, and we shouldn't we have: $$ \min f\leq \int_{[0,1]^2}f dA\leq \max f $$ While, if we integrate along the curve $\gamma_n$, we should have $$ \text{length}(\gamma)\min f\leq\int_{\gamma_n}fdt\leq\text{length}(\gamma)\max f $$ Then as the length of the curve goes to infinity as $n$ goes to infinity then we should get that $\lim_{n\rightarrow\infty}\int_{\gamma_n}fdt\rightarrow \infty$. Thus, it seems that my initial intuition was wrong. However, I may be completely wrong and they might be equal (my intuition isn't always the best). Either way, can someone confirm that these two integrals are not equal to each other, or if they actually are (or if there is a setting where it would make sense for them to be equal) I would be interested in a reference? Thanks.