$\tilde{P}$ is a refinement of $P$. $m_j\le \tilde{m_p}$? $\tilde{P}$ is a refinement of $P$. $P=\{x_0,....,x_n\}$ and $\tilde{P}=\{x_{k_0},...,x_{k_n}\}$, and $x_{k_j}=x_j.$  $m_j = \inf\{f(x):x_{j-1}\le x \le x_j\}$, and $\tilde{m_p}=\inf\{f(x_p):x_{k_{j-1}}\le x_p \le x_{k_j}\}$.  My textbook says $m_j\le \tilde{m_p}$, but shouldn't it be the opposite? because $P \subset \tilde {P}$.
Edit: I add the photo in case the above explanation is not enough. 


Let's start with the observation that if $I$ is an interval, and $I' \subset I$ is a subinterval, then $\inf \{f(x) : x \in I\} \leq \inf \{f(x) : x \in I'\}$, since the $\inf$ on the whole interval is clearly smaller than the $\inf$ on the contained interval.
Now let 
$$ m_j = \inf\{f(x):x_{j-1}\le x \le x_j\}$$.
If we refine this interval, we get a bunch of points 
$$\tilde x_{0} \  \tilde x_{1} \ ... \ \tilde x_{m}$$
with $\tilde x_0 = x_{j-1}$ and $\tilde x_{m} = x_j$
Now, for any $p < m$, we have 
$$(\tilde x_p, \tilde x_{p+1}) \subset (x_{j-1}, x_j) $$ 
And so
$$\inf \{f(x) : x \in (x_{j-1}, x_j) \} \leq \inf \{f(x) : x \in x_{p}, x_{p+1} \} $$
