# Proving a Riemann integral theorem

Question: Prove that for $f$ continuous on $[a,b] \in \Bbb R$, for $c \in [a,b]$, then: $$\int_a^b f(x)\, dx = \int_a^c f(x)\,dx + \int_c^b f(x)\, dx$$

My Work:

Let $\{P_k\}$ be a sequence of partitions such that the mesh of $P$ goes to $0$ as $k \to \infty$. For each $k$, denote the points of the partition $P_k$ by $x_0^k < x_1^k < \cdots < x_N^k$. In each subinterval $[x_{i-1}^k, x_i^k]$ choose a point $x_i^{k*}$.

W.L.O.G. let $c \in [a,b]$ be an endpoint of one of the subintervals. If $c$ is not an endpoint, we can refine the partition a point $x_j^k$ at $c$. Then we have: $$\int_a^c f(x)\,dx = \sum_{i=1}^j f(x_i^{k*})(x_i^k - x_{i-1}^k) \quad\text{and}\quad \int_c^b f(x)\, dx = \sum_{i=j+1}^N f(x_i^{k*})(x_i^k - x_{i-1}^k)$$ Adding the two partial sums together, we have $$\sum_{i=1}^N f(x_i^{k*})(x_i^k - x_{i-1}^k) =\int_a^b f(x)\, dx$$

Needing Work: I am not sure I can make the WLOG argument I made, but I'm not sure how else to introduce $c$. And it just seems a little lacking.

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Yes, this is the way to reason. With the Riemann integral you can always refine the partition to arrive at a valid partition. One can show that refining the partition improves on the approximation. This then easily gives you a reason for the wlog. – user20266 Aug 18 '12 at 15:04

Just note that $f$ is continuous, so that the sum converges with respect to refinement of partion. That is, you can add a point $c$ to your partition, this will be closer to the limit state, and the limit(ie, the integral) will not change.