Understanding the proof for Lebesgue integrability criterion I found a somewhat rigorous proof of Lebesgue's integrability criterion, where $\Delta x_k = x_k - x_{k-1}$, and
$$M_k = \sup_{x \in [x_{k-1},x_k]} f(x),m_k = \inf_{x \in [x_{k-1},x_k]} f(x)
$$
However, how is the proposition
$$ D_i \subseteq \bigcup_{k \in T} [x_{k-1},x_k]
$$
justified?
We read "$T$ is the set of those $k$ with $D_i\cap (x_{k-1},x_k)\neq \emptyset$."  Furthermore, the $[x_{k-1},x_k]$ intervals collectively cover $[a,b]$ per the statement that $a = x_0&lt;x_1&lt; \cdots &lt; x_n=b$.  We are also told that $[a,b]$ is the domain of $f$, and $\omega_f$ is defined on that domain of definition, so $D_i$, defined as the set of points $x$ where $\omega_f(x)$ exceeds a certain value, must be a subset of that domain.  

We thus have $D_i\subset \bigcup_{k=1}^n [x_{k-1},x_k]$.  From the latter union, we may omit all those $[x_{k-1},x_k]$ where $D_i$ does not intersect $(x_{k-1},x_k)$ without affecting the correctness of the inclusion, a conclusion which is along the lines of "if $X\subset A\cup B\cup C$ and $X\cap C=\emptyset$, then $X\subset A\cup B$."  By the definition of $T$, those are precisely the intervals which are omitted in the union $\bigcup_{k\in T} [x_{k-1},x_k]$.  We thus have $D_i \subseteq \bigcup_{k \in T} [x_{k-1},x_k]$ as desired.
We read "$T$ is the set of those $k$ with $D_i\cap (x_{k-1},x_k)\neq \emptyset$."  Furthermore, the $[x_{k-1},x_k]$ intervals collectively cover $[a,b]$ per the statement that $a = x_0&lt;x_1&lt; \cdots &lt; x_n=b$.  We are also told that $[a,b]$ is the domain of $f$, and $\omega_f$ is defined on that domain of definition, so $D_i$, defined as the set of points $x$ where $\omega_f(x)$ exceeds a certain value, must be a subset of that domain.  

We thus have $D_i\subset \bigcup_{k=1}^n [x_{k-1},x_k]$.  From the latter union, we may omit all those $[x_{k-1},x_k]$ where $D_i$ does not intersect $(x_{k-1},x_k)$ without affecting the correctness of the inclusion, a conclusion which is along the lines of "if $X\subset A\cup B\cup C$ and $X\cap C=\emptyset$, then $X\subset A\cup B$."  By the definition of $T$, those are precisely the intervals which are omitted in the union $\bigcup_{k\in T} [x_{k-1},x_k]$.  We thus have $D_i \subseteq \bigcup_{k \in T} [x_{k-1},x_k]$ as desired.