How to prove $L_{f}(P) \leq L_{f}(Q)$ when $Q$ and $P$ are partitions of $[a,b]$ and $Q \supseteq P$ I'm having trouble proving this idea.
Suppose that $f$ is bounded on the interval $[a, b]$. $P$ and $Q$ are partitions of $[a, b]$, and $Q \supseteq P$.
$$
L_{f}(P) \leq L_{f}(Q)
$$
I know that this makes sense, because by adding points to a partition, the subintervals get smaller which makes the minima $m_i$ larger, therefore making the lower sums bigger. But I'm not sure if this is a valid proof.
 A: First, note we can simply prove this for $P$ and $Q$ where $Q$ is obtained from $P$ by adding only one number $t^*$, and then go by induction on the number of extra points $Q$ has. 
So suppose $P=\{t_0,\ldots,t_i,t_{i+1},\ldots,t_n\}$ and $Q=\{t_0,\ldots,t_i,t^*,t_{i+1},\ldots,t_n\}$. Let $m_i$ be the infimum of $f$ on $[t_i,t_{i+1}]$ and $m',m''$ the infimum of $f$ on $[t_i,t^*]$ and $[t^*,t_{i+1}]$ respectively. Then $L_f(Q)$ and $L_f(P)$ differ by $$m''(t_{i+1}-t^*)+m'(t^*-t_i)-m_i(t_{i+1}-t_i)$$
Can you show this is nonnegative? Remember that if $A\subseteq B$ then $\inf A\geqslant \inf B$. 
A: Let
$$m_k = \inf \{f(x): x_{k-1} \leqslant x \leqslant x_k\}.$$
If you add another partition point $\xi_k \in (x_{k-1}, x_k)$, then
$$m_k' = \inf \{f(x): x_{k-1} \leqslant x \leqslant \xi_k\} \geqslant m_k, \\
m_k'' = \inf \{f(x): \xi_k \leqslant x \leqslant x_k \} \geqslant m_k,$$
and
$$m_k'(\xi_k-x_{k-1}) + m_k''(x_k-\xi_k)\geqslant m_k(x_k-x_{k-1}).$$
Use this argument to show that
$$L_f(Q) \geqslant L_f(P).$$
