If you consider the integral from a to b on h(x) then that would be equal to the integrals of a to c, added to c to b. This is quite straight forward, and let this one differing point be c.
Then
Integral from a to b of h(x)
= Integral from a to c-dx of h(x) added to integral from c+dx to b of h(x)
[as dx -> 0]
= integral from a to b of g(x)
Where g(x) will then only differ at the point c.
That's pretty much a hand waving arguement I think though.
Actually it can be discontinuous a set of measure zero.
But I suspect the person who set this question expected something much less powerful. Here is a quick proof of what I think is meant by this problem.
Suppose that z in an interior point of (a,b). There is a e>0 such that (z-e,z+e) is a subset of [a,b]. Now by additively g is integrable on each of these subintervals: [a,z-e], [z-e,z+e] and [z+e,b]. Only on [z-e,z+e] have you changed the definition g so we can control the variation that interval, after all its length is only 2e. If we make e small enough the sum remains the same.
Keep at it. Sometimes you just get to a point where you need to be shown how to do many things, but you get past it at some stage. Once you get to a point where you've had enough "exposure" you'll be able to do all the things someone once had to show you and then YOU can "wow" the more inexperienced students!
-Dan
No, for example, you where able to understand about what I said about rational discontinuities. And furthermore were amazed by the statement. If you were at the limit you would not have understood what I said nor would have found the statement amazing. When you start to reach a point in math when you are amazed by some statements (for example, there exists a function which is continous everywhere and differenciable nowhere!!) then you are going good.