Refinement for Lebesgue Integral 
Let $f$ be measurable on $[a,b]$ and let $P_1$ and $P_2$ be measurable partitions of $[a,b]$. If $P_2$ is a refinement of $P_1$, show that $$L[f,P_1]\leq L[f,P_2]\quad\text{and}\quad U[f,P_2]\leq U[f,P_1]$$

Definition:
(1) Let $P=\{E_j\}_{j=1}^n$ and $P^*=\{F_k\}_{k=1}^m$ be two measurable partitions of $[a,b]$. We say $P^*$ is a refinement of $P$ if for every $k$, there is a $j$ such that $F_k\subseteq E_j$
(2) The upper sum $U[f,P]=\sum_{j=1}^nM_j \underbrace{m(E_j)}_{\text{Lebesgue Measure}}$ where $M_j=\sup_{x\in E_j}f(x)$
(3) The lower sum $L[f,P]=\sum_{j=1}^nm_j \underbrace{m(E_j)}_{\text{Lebesgue Measure}}$ where $m_j=\inf_{x\in E_j}f(x)$

For Riemann Integral, we can split the interval and use the Infimum property.
But for Lebesgue Integral, I couldn't understand how to write things in rigorous way? As, we use Lebesgue Measure for the set. How can we use the fact $F_k\subseteq E_j$ to trace the sum? Because here, $$F_k\subseteq E_j \implies m(F_k)\leq m(E_j)$$And infimum/supremum is not less/greater than before. It will be a great help for me if anyone gives me any hint or answer to demystify this.
 A: I give an answer which assumes that the properties of Lebesgue integral are already known, at least for simple functions, namely for finite linear combinations of indicator functions. Otherwise, the proof is a bit longer.
According to the definitions above, $L(f,P)$ is the integral of the function $$f_P := \sum_{j=1}^n \big( \inf_{E_j} f \big) \mathbb1_{E_j}$$
with regard to the Lebesgue measure. Hence it suffices to check that when $P^*$ is a refinement of $P$, one has $f_P \le f_{P^*}$ everywhere on $[a,b]$.
To check this, fix $x \in [a,b]$, call $E_j$ (resp $F_k$) the only block of $P$ (resp $P^*$) which contains $x$. Then $F_k \subset E_j$, so $f_P(x) = \inf_{E_j} f \le \inf_{F_k} f = f_{P^*}(x)$.
I now complete the proof for the case where the properties of Lebesgue integral of simple functions are not known yet.
For each $j \in [1,n]$, choose $a_j$ in $E_j$ and let $K_j = \{k : F_k \subset E_j\}$. Since $P^*$ is a refinement of $P$, each $E_j$ is the disjoint union of the $F_k$ over all $k \in K_j$ (proof below). For each index $k$, choose $b_k \in F_k$. Then
\begin{eqnarray*}
L(f,P) 
&=& \sum_j f_P(a_j) m(E_j) \\
&=& \sum_j f_P(a_j) \sum_{k \in K_j}m(F_k) \\
&=& \sum_j \sum_{k \in K_j} f_P(b_k) m(F_k) \textrm{ since $f_P$ is constant on each $E_j$} \\
&=& \sum_k f_P(b_k) m(F_k) \\
&\le& \sum_k f_{P^*}(b_k) m(F_k) \textrm{ since } f_P \le f_{P^*} \\
&=& L(f,P^*).
\end{eqnarray*}
Proof that each $E_j$ is the (disjoint) union of the $F_k$ over all $k \in K_j$.
We say $P^*$ is a refinement of $P$ if for every $k$, there is a $j$ such that $F_k \subset E_j$. Such a $j$ is unique because $P$ and $P^*$ are partitions.
Given $x \in E_j$, since $P^*$ is a partition on $[a,b]$, $x$ belongs to some $F_k$. By assumption, $F_k$ is contained in some $E_{j'}$. Since the subsets $E_{j'}$ are pairwise disjoint, $j'=j$, hence $k \in K_j$.
Conversely, if $x \in F_k$ for some $k \in K_j$, then $x \in E_j$ since $F_k \subset E_j$.
A: Let me give you some hints:
From your definition for every $k$, there is a $j$ such that $F_k\subseteq E_j$. Could you trace $F_k$'s for fixed $E_i$. Construct a set $A_i$.

 $A_i=\{F_\lambda:F_\lambda\subset E_i\}$

Could you show that $m(E_i)=\sum_{F_\lambda \in A_i}m(F_{\lambda})$?

 To prove $m(E_i)\leq\sum_{F_\lambda \in A_i}m(F_{\lambda})$ and $m(E_i)\geq\sum_{F_\lambda \in A_i}m(F_{\lambda})$ use $E_i=\bigcup F_j\cap E_i$ and $F_k'=F_k\setminus\left(\cup_{i=1}^{k-1} F_{\lambda_j}\right)$ (construct a pairwise disjoint partition) respectively.

Now from $U[f,P_1]=\sum_{i=1}^n\sup_{x\in E_j}f(x)\:m(E_i)$ try to show $U[f,P_1]\geq U[f,P_2]$.

 Use $m(E_i)=\sum_{F_\lambda \in A_i}m(F_{\lambda})$ and $\sup_{x\in E_i} f(x)>\sup_{x\in F_{\lambda}} f(x)$. You might be encounter double summation, but you can carefully handle this and turn them into a single summation.

I might be sloppy with some notation. If you need any further clarification, then ask me on comment. Cheers.
