Note that A is [2 2]
A = size([0 0; 1 1])
so size(A) is [1 2]
size(A)
So when you compare size(Out(dates).aerosol) with size(A) here:
if size(Out(dates).aerosol)>=size(A)
you are checking size(Out(dates).aerosol) >= [1 2], and since size(Out(dates).aerosol) can be [1 2], [2 2] or [4 2], that comparison is always [true true]
size([0 0]) >= size(A)
size([0 0; 0 0]) >= size(A)
size([0 0; 0 0; 0 0; 0 0]) >= size(A)
so the code inside the if block is always executed, and element (2,2) of Out(dates).aerosol is always (attempted to be) retrieved
SM_p029r029(dates).Aerosol1 = Out(dates).aerosol(2,2)
but in case size(Out(dates).aerosol) is [1 2], that's an error because element (2,2) doesn't exist.
What you really mean to be doing is, instead of comparing to size(A), compare to A, since A is [2 2]
size([0 0]) >= A
size([0 0; 0 0]) >= A
size([0 0; 0 0; 0 0; 0 0]) >= A
That way, the else block will be executed when size(Out(dates).aerosol) is [1 2]. (A vector if condition is considered true only if all elements of the condition are true.)
However, you can do away with the if/else entirely and just do:
SM_p029r029(dates).Aerosol1 = Out(dates).aerosol(min(size(Out(dates).aerosol,1),2),2)
because the expression min(size(Out(dates).aerosol,1),2) gives you either the index of the last row of Out(dates).aerosol or 2, whichever is smaller, which is exactly what you need
min(size([0 0],1),2)
min(size([0 0; 0 0],1),2)
min(size([0 0; 0 0; 0 0; 0 0],1),2)
Or more succinctly:
SM_p029r029(dates).Aerosol1 = Out(dates).aerosol(min(end,2),2);
