Sum of Integrals on Adjacent Intervals for Integrable Functions/Lemma/Proof 2

Proof
Let $P = \set {x_0, x_1, \ldots, x_n}$.

Suppose that:
 * $c \in P$

Then:
 * $Q = P$

We have:
 * $\map L P \ge \map L P$
 * $\leadsto \map L Q \ge \map L P$ as $Q = P$

This finishes the proof for this case.

The only other possibility for $c$ is:
 * $x_{j-1} < c < x_j$

where $1 \le j \le n$.

Let $m_i$ be the infimum of $f$ on the interval $\closedint {x_{i - 1}} {x_i}$ for all $i \in \set {1, 2, \ldots, n}$.

We have:

Let $r$ be the infimum of $f$ on the interval $\closedint {x_{j - 1}} c$.

Let $s$ be the infimum of $f$ on the interval $\closedint c {x_j}$.

The interval $\closedint {x_{j - 1}} c$ is a subset of $\closedint {x_{j - 1}} {x_j}$.

Therefore, a lower bound of $f$ on the interval $\closedint {x_{j - 1}} {x_j}$ is a lower bound of $f$ on the interval $\closedint {x_{j - 1}} c$ as well.

The infimum of $f$ on the interval $\closedint {x_{j - 1}} {x_j}$ is a lower bound for $f$ on this interval.

Accordingly, the infimum of $f$ on the interval $\closedint {x_{j - 1}} {x_j}$ is a lower bound of $f$ on the interval $\closedint {x_{j - 1}} c$ as well.

Therefore, the infimum of $f$ on the interval $\closedint {x_{j - 1}} c$ is greater than or equal to the infimum of $f$ on the interval $\closedint {x_{j - 1}} {x_j}$.

In other words:
 * $r \ge m_j$

A similar analysis of $\closedint c {x_j}$ gives:
 * $s \ge m_j$

We find:

This finishes the proof.