$f$ is integrable $\iff$ for every sub-block $B$ we have that the function $f|_{B}$ is integrable, i.e. $\int_{A}f=\sum_{B}\int_{B}f|_{B}$ https://ift.tt/eA8V8J

QUESTION: Let $f:A \rightarrow \mathbb{R}$ be a limited function and let $P$ be a partition of the block $A$ ($A$ is a block in $\mathbb{R}^m$). Then $f$ is integrable $\iff$ for every sub-block $B$ we have that the function $f|_{B}$ is integrable and in this case, $$\int_{A}f=\sum_{B}\int_{B}f|_{B}$$.

REMARK: The professor allowed us to use the following concepts:

1. Proposition: Let $P_0$ be an arbitrary partition of the block $A$. In order to consider the upper and lower integrals of the limited function $f:A \rightarrow \mathbb{R}$, we just need to consider partition refinements of $P_0$. That is, we have $$\underline\int_{A} f(x) dx= \underset{P\supset P_0}{sup} s(f; P)$$ and $$\overline\int_{A} f(x) dx= \underset{P\supset P_0}{inf} S(f; P)$$
2. Theorem: The limited function $f: A \rightarrow \mathbb{R}$ is integrable $\iff$ for every $\epsilon>0$ it is possible to find a partition $P$ of the block $A$ such that $$\displaystyle\sum_{B\in P} \omega_{B}\cdot vol B<\epsilon$$ Where $\omega_{B}$ is the set of the oscillations, i. e., $$\omega_{B}:= sup\{|f(x)-f(y)|; x, y \in B\}$$

MY ATTEMPTY:

$(\Longrightarrow)$ Let $f: A \rightarrow \mathbb{R}$ be a limited function and let $P$ be a partition of the block $A$. Suppose that $f$ is integrable then $\forall \epsilon >0$ it is possible to obtain an partition $P=P_1 \times \cdots \times P_n$ of $A$ such that $\displaystyle\sum_{B\in P} \omega_B \cdot \text{vol}B <\epsilon$, where $B$ are blocks in $P$. Once $B$ are sub-blocks of $A$, let $P_0$ be an partition of $B$. Therefore for every limited function $f|_{B}$ we just need to consider the refinement partitions of $P_0$. Indeed, let $B=\displaystyle\Pi_{i=1}^{n}[b_i, c_i] \subset A$ then for every $i=1, \cdots, n$ lets define $Q_i= P_i\cap[b_i, c_i]$ from this we have a new partition $Q = Q_1 \times \cdots \times Q_n$ of $A$ that is a refinement of $P$ and, furthermore, the blocks of $Q$ are contained in $B$ makes a partition $P_0$ of $B$. Thus $$\underbrace{\displaystyle\sum_{B'\in P_0}\omega_{B'}\cdot \text{vol} B'}_{(I)}\leq\displaystyle\underbrace{\sum_{B\in P}\omega_{B}\cdot \text{vol} B<\epsilon}_{(II)}$$ $(I) \subset (II)$ therefore $f|_{B}$ is intagrable.

$(\Longleftarrow)$ We just need to consider $P=P_1 \times \cdots \times P_n$ as a partition of the block $A$ and we also need to consider that this partition is a composition of the block $A$ in sub-blocks like $B=I_1 \times \cdots \times I_n$ where every $I_j$ is an interval of the partition $P_j$, where every sub-block $B$ is the block of partition $P$, i.e., $B\in P$. So, writting $A=\displaystyle\bigcup_{i=1}^{n}B_i$ and remembering that every $f|_{B}$ is integrable. Note that if $P_i$ is a partition of $B_i$ we can consider $Q=\displaystyle\sum_{i=1}^{n}P_i$ as an refinement partition of $P$ thus $f:A \rightarrow\mathbb{R}$ is integrable.

Now we just need to show that: $$\int_{A} f \leq \displaystyle\sum_{B \in P} \int_{B} f|_{B}$$.

In $f:A \rightarrow \mathbb{R}$ considering the partition $P$ of the block $A$ we just need to consider refinement partitions of $P$, let $Q$ be an arbitrary partition of the block $A$ we can consider, for instance $P_0= P+Q$. It follows from upper integration definition that $$s(f, P)=\displaystyle\sum_{B \in P} m_B(f)\cdot \textbf{vol}B= \displaystyle\sum_{B \in P} m_{B}(f|_{B}) \cdot \textbf{vol} B$$. Then, for every $B$ we consider $B' \subset B$, the sub-blocks of $B$ resultants of the refinement of $P$, and $B=\bigcup B'$. Therefore, \begin{align*} \int_{A} f = \displaystyle sup_{P_0\supset P} s(f, P_0)& = sup \left(\displaystyle\sum_{B\in P}m_{B}(f|_{B}) \cdot \textbf{vol} B\right)\\ & = sup \left(\displaystyle\sum_{B\in P}m_{B}(f|_{B}) \displaystyle\sum_{B'\subset B} \textbf{vol} B'\right)\\ & = sup \left(\displaystyle\sum_{B\in P}\displaystyle\sum_{B'\subset B}m_{B}(f|_{B}) \textbf{vol} B'\right)\\ & \leq sup \left(\displaystyle\sum_{B\in P}\displaystyle\sum_{B'\subset B}m_{B'}(f|_{B}) \textbf{vol} B'\right)\\ & = \displaystyle\sum_{B\in P} sup \left(\displaystyle\sum_{B'\subset B}m_{B'}(f|_{B}) \textbf{vol} B'\right)\\ & = \displaystyle\sum_{B\in P} \underline{\int_{B}} f|_{B}\\ & = \displaystyle\sum_{B\in P} \int_{B} f|_{B} \end{align*} Thus, $$\int_{A} f \leq \displaystyle\sum_{B\in P} \int_{B} f|_{B}$$

Similarly, we can show for upper sum, and obtain $$\int_{A} f \geq \displaystyle\sum_{B\in P} \int_{B} f|_{B}$$ And finally, conclude $$\int_{A} f = \displaystyle\sum_{B\in P} \int_{B} f|_{B}$$

MY DOUBT: Would you help me to improve my answer? Specialy in this $(\Longleftarrow)$ way.

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