# 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!

## 1 Answer

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 $h \to 0$.....