Assume g is integrable over an interval, say [0,1]. Let f(x) be the integral of g from 0 to x. Let s be a point inside the interval. Remember that the integral from 0 to s+h is the integral from 0 to s plus the integral from s to s+h. Assuming g is bounded, the latter integral goes to 0 as h approaches 0. Thus f is continuous at s. It doesn't matter how jumpy g is, f is continuous.
Next assume g is continuous at s. The difference quotient becomes the integral of g from s to s+h, divided by h, as h approaches 0. Since g is continuous at s, we can restrict g to within ε of g(s), which places the same restriction on the integral divided by its width. The latter is our difference quotient. Therefore g is the derivative of f. Use one-sided limits when s is one of the endpoints.
Here is the intuition behind the theorem. Let a vertical line bound the area under the curve, and slide the vertical line steadily to the right. How is the area increasing with time? The amout of area added is the height of the line segment, or g(s).
Next assume h is differentiable over an interval, with derivative g. Let f be the integral of g, as described above. Now h′ = f′ = g. Subtract to show the derivative of f-h is 0; thus f-h is a constant. The integral of the derivative is a constant away from the original function. Integration and differentiation are indeed inverse operations.
We don't have to compute lower sums any more, as we did in our earlier examples. The area under the parabola y = x2, from 0 to x, is x3/3, because the derivative of x3/3 is x2.