Let f be an integrable function on a region S in the plane. (The reasoning works in n dimensions, but it's easier to picture in the plane.) Let g be a 1-1 function that maps the region T onto the region S. Thus g has two input variables and two output variables, and g is invertible. For example, g could be a linear transformation that shifts or rotates the plane. Or g could be much more complicated.
Assume g is differentiable over T, and furthermore, the derivative, which is represented by the jacobian matrix, has a positive determinant throughout T.
Displaying perfect foreknowledge, let's evaluate the integral of f(g)×det(jac(g)) over T.
Choose a Riemann net over T with granularity δ, where δ is small. Let p be the lower left corner of one of the rectangles in the Riemann net. Let q = g(p), a point in S. Move to the right of p, along the x axis, and draw the image of this line segment in S. The image starts at q and proceeds in a direction that is given by the partial of g with respect to x. When we are moving parallel to a coordinate, the partial derivative is the directional derivative. This is the first row of the jacobian. Next, start at p and move up, parallel to the y axis. The image starts at q and moves in the direction given by the partial of g with respect to y, i.e. the second row of the jacobian. The two image segments need not meet at a right angle. In fact they could meet at any angle at all, other than 0° and 180°. If the two rows of the jacobian are parallel, or antiparallel, the jacobian has a zero determinant, and we've ruled that out.
The rectangle based at p maps onto a cell in S, based at q. The cell is approximately a parallelogram. We've already drawn two sides of this parallelogram, adjacent to q. Now draw the other two sides, corresponding to the other two sides of the rectangle in T. But is the image really a parallelogram? As soon as we leave the vicinity of q, g could become very strange.
This is where we need the continuity of the partial derivatives. Since T is closed and bounded, a continuous function is uniformly continuous. Make δ small enough, so that the partials don't very by more than ε within any given rectangle. Now the image of the bottom of our rectangle can't turn sharply or wiggle all about. It might curve a bit, but it is close to the line that starts at q and moves in the direction given by the partial at p. This can be made explicit, as a function of ε. This holds for all four sides of the parallelogram. Therefore the image of the rectangle in T is very close to the parallelogram based at q, whose sides are determined by the jacobian of g at p.
The area of a parallelogram in the plane is given by the determinant of its spanning vectors. In this case the two sides of the parallelogram are the two partials times the width and height of the original rectangle. This is the determinant of the jacobian times the area of the rectangle. The determinant of the jacobian is a local magnification factor on area. Multiplying f(g(p)) times the determinant of the jacobian at p is practically the same as multiplying f(q) times the area of the cell based at q, which is the image of the original rectangle. This is starting to look like the integral of f over S.
At this point we need the previous theorem. The cells in S aren't rectangles. In fact they look more like parallelograms. But it doesn't matter. All nets lead to the integral of f on S, provided the granularity drops to zero.
what is the diameter of one of these cells? Remember that the image of a rectangle is approximately a parallelogram. Its diameter, from one corner to the other, is no worse than the sum of the two segments adjacent to q. This happens when the two segments are almost parallel, a very pointy parallelogram. How long are these segments? The base of the rectangle is multiplied by the length of the partial with respect to x, and the height of the rectangle is multiplied by the length of the partial with respect to y. These partials are continuous on a closed bounded region, namely T, hence there is an upper bound. Let b be an upper bound. Now the sides of the image parallelogram are no worse than b times the sides of the rectangle. The cells in S have granularity 2bδ, and that shrinks to zero as δ goes to zero.
The Riemann nets on T map to general nets on S, and the Riemann sum on T is arbitrarily close to the sum over the corresponding general net on S. I'm glossing over a lot of ε δ mathematics here, but a detailed proof would take several pages, and I think you get the idea. The function g carries tiny rectangles to tiny parallelograms, and the area of each rectangle is multiplied by the determinant of the jacobian at that point. Therefore the integral of f over S is the same as the integral of f(g) times the determinant of the jacobian of g, evaluated over T.
This proof generalizes to higher dimensions.