I have been looking for uncountable large structures that generalize the idea of matrix multiplication and noticed the following pattern
Given functions $M_1(x,y) , M_2(x,y)$ defined on the unit square $[0,1] \times [0,1]$ one can define their "matrix product" as
$$ M_3(x,y) = \int_{0}^{1} M_1 (s, y) M_2 (x, 1-s) ds = ( M_1 M_2 )$$
If we try to approximate this integral with riemann sums, we find that the approximations are computationally equivalent to multiplying ever higher order matrices (to produce a third matrix), but it's not clear that associativity is preserved as we take the limit to forming an integral.
My work:
for 3 functions A,B,C I expressed $A(BC)$ and $(AB)C$ in terms of nested integrals but now am not clear on how to show that the 2 expressions are equal. One idea was to perturb the definition above to: $ \int_{-\infty}^{\infty} M_1 (s, y) M_2 (x, -s) ds$ and then to show associativity for this simpler looking operation, [it then is easy to show that maps sending the open unit square to the whole plane, composed with this operation, and then said map's inverse is equivalent to the earlier operation]
Let $A,B,C:[0,1]^2\to \mathbb R$ be matrices. $$ \begin{align} ((AB)C)(x,y) &= \int_0^1 (AB)(s,y)C(x,1-s)\,ds \\&=\int_0^1\left(\int_0^1A(t,y)B(s,1-t)\,dt\right)C(x,1-s)\,ds \\&=\int_0^1\int_0^1A(t,y)B(s,1-t)C(x,1-s)\,dt\,ds \end{align}$$$$ \begin{align} (A(BC))(x,y) &= \int_0^1 A(s,y)(BC)(x,1-s)\,ds \\&=\int_0^1A(s,y)\left(\int_0^1B(t,1-s)C(x,1-t)\,dt\right)\,ds \\&=\int_0^1\int_0^1A(s,y)B(t,1-s)C(x,1-t)\,dt\,ds \\&\stackrel{\text{Fubini}}=\int_0^1\int_0^1A(s,y)B(t,1-s)C(x,1-t)\,ds\,dt \\&\stackrel{s\leftrightarrow t}=\int_0^1\int_0^1A(t,y)B(s,1-t)C(x,1-s)\,dt\,ds \\&= ((AB)C)(x,y) \end{align} $$