Let $f: [a, b]\longrightarrow \mathbb R$ be a bounded function and let $\mathcal{P}=\{a=x_0<\ldots<x_n=b\}$ be a partition of $[a, b]$.
The lower Riemann sum of $f$ with respect to $\mathcal{P}$ is
$$L(f, \mathcal{P})=\displaystyle \sum_{j=1}^n (x_{j+1}-x_j)\inf_{[x_{j}, x_{j+1}]} f.$$
I'd like to show that if $\mathcal{P}\subset \mathcal{P}^\prime$ are partitions of $[a, b]$, then
$$L(f, \mathcal{P})\leq L(f, \mathcal{P}^\prime).$$
Attempt: The proof will be done via induction on the cardinality of the set $\mathcal{P}^\prime\setminus \mathcal{P}$. Suppose $\mathcal{P}^\prime\setminus\mathcal{P}$ contains a single element, let us say $a$. This allows us to write
$$\mathcal{P}^\prime=\{x_0, \ldots, x_k, a, x_{k+1}, \ldots, x_n\}.$$ But $x_k<a<x_{k+1}$ implies \begin{align*} [x_k, a]\subset [x_k, x_{k+1}]\quad \textrm{e}\quad [a, x_{k+1}]\subset [x_k, x_{k+1}] \end{align*} and therefore \begin{align*} \inf_{[x_k, x_{k+1}]} f\leq \inf_{[x_k, a]} f\quad \textrm{and}\quad \inf_{[x_{k}, x_{k+1}]} f \inf_{[a, x_{k+1}]} f. \end{align*} Hence: \begin{align*} (x_{k+1}-x_k) \inf_{[x_k, x_{k+1}]} f&= (x_{k+1}-a) \inf_{[x_{k}, x_{k+1}]} f+(a-x_k) \inf_{[x_k, x_{k+1}]} f\\ &\leq (x_{k+1}-a) \inf_{[a, x_{k+1}]} f+ (a-x_k) \inf_{[x_k, a]} f. \end{align*} Using this, it follows: \begin{align*} L(f, \mathcal{P})&=\sum_{j=1}^n (x_{j+1}-x_j) \inf_{[x_j, x_{j+1}]} f\\ &=\sum_{j=1}^{k-1} (x_{j+1}-x_j) \inf_{[x_j, x_{j+1}]} f+(x_{k+1}-x_k) \inf_{[x_k, x_{k+1}]} f+\sum_{j=k+1}^n (x_{j+1}-x_j)\inf_{[x_j, x_{j+1}]} f\\ &\leq \sum_{j=1}^{k-1} (x_{j+1}-x_j) \inf_{[x_j, x_{j+1}]} f+(x_{k+1}-a) \inf_{[a, x_{k+1}]} f+ (a-x_k) \inf_{[x_k, a]} f+\sum_{j=k+1}^n (x_{j+1}-x_j)\inf_{[x_j, x_{j+1}]} f\\ &=L(f, \mathcal{P}^\prime). \end{align*} This proves the base step of the induction. Can someone help me finish the induction?
Thanks.
P.s.: Maybe I shouldn't use induction? For, if $\mathcal{P}^\prime\setminus \mathcal{P}$ contains $l+1$ elements, then there are partitions $\mathcal{P}_0=\mathcal{P}, \ldots, \mathcal{P}_l=\mathcal{P}^\prime$ such that $\mathcal{P}_0\subset \ldots\subset \mathcal{P}_{l+1}$ and $\mathcal{P}_{l+1}\setminus \mathcal{P}_l$ has a single element, and therefore:
$$L(f, \mathcal{P})=L(f, \mathcal{P}_0)\leq L(f, \mathcal{P}_1)\leq \ldots L(f, \mathcal{P}_{l+1})=L(f, \mathcal{P}^\prime).$$
Maybe I messed up the indices, but is the idea right?
The base case is right (maybe apart from some typos which I may have missed). The main idea is that if you have two bounded sets $A\subset B$ then $\inf B \leq \inf A$: "infimum of a smaller set is larger" (and similarly "supremum of smaller set is smaller" if you want to prove an analogous statement for upper sums).
Finally, induction is a very good idea, and what you wrote in the end is a correct way of proving things.
Rewriting the last part formally:
Let $f, [a,b]$ be as in the problem statement. Denote $\Omega$ to be the set of $l\in \Bbb{N}$ (here $\Bbb{N} = \{1,2,3,\dots\}$ doesn't include $0$) such that for every pair of partitions $P$ and $Q$ of $[a,b]$ with $P\subset Q$ and $|Q\setminus P| = l$, we have $L(f,P) \leq L(f,Q)$. We are going to show $\Omega = \Bbb{N}$ using induction.
Based on the first part of the proof, we have proven that $1\in \Omega$ (i.e the base case is done). Now, suppose $l\in \Omega$; we'll show $l+1\in \Omega$. Let $P,Q$ be any partitions of $[a,b]$ with $P\subset Q$ and $|Q\setminus P| = l+1$. Fix a point $\xi \in Q\setminus P$, and consider the set $R = Q\setminus\{\xi\}$. Then, the following assertions are easily verified:
Therefore: \begin{align} L(f,P) \leq L(f,R) \leq L(f,Q), \end{align} where the first inequality is by the induction hypothesis (which can be used since $|R\setminus P| = l$) and the second is by the base case (which is valid since $|Q\setminus R| = 1$).
This shows $l+1\in \Omega$. Therefore, by the principle of induction, $\Omega = \Bbb{N}$, and this is precisely what was to be proven.