# Proving that if $f,g$ differ at a finite number of points then the lower Riemann integrals are equal

Let $f,g : [a,b] \to R$ be bounded and suppose that $f(x)$ not equal to $g(x)$ only at a finite number of points.

Prove that the lower Riemann integrals of $f$ and $g$ are equal to each other, and prove the same case with the upper integrals.

I have been told to use the fact that $L(f,P)$ tends to the lower integral given that the mesh of partition tends to $0$ (same for upper integral).

I've been thinking of expanding $L(f,P) - L(g,P)$ and commenting on how the infimums will be equal for many intervals and showing that that tends to $0$ as the mesh of $P$ tends to $0$ and therefore the two limits equal each other, implying the lower Riemann integrals are equal.

Need some help with the details however!

Because the mesh gets finer and finer, you might as well start with a mesh so fine that each point of difference is in its own interval. You can do this because with a finite set of points there is a minimum distance between any pair. Now let there be $n$ points where $f,g$ disagree. If the mesh spacing is $h$, the difference in Riemann sums is bounded by $h$ times the sum of the absolute value of the differences. As $h \to 0$.....