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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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    $\begingroup$ You haven't made it clear form the start that the $B's$ are nonoverlapping and $A = \cup_{B \subset A} B$, although it is implicit in the argument. Why have you used the lebesgue-integral tag but based your argument on Riemann-Darboux sums? I assume then that "integrable" here can be taken as "Riemann integrable" in what you are trying to prove. $\endgroup$
    – RRL
    Nov 11, 2020 at 19:24
  • $\begingroup$ Yes, you're write isn't lebesgue-integral. $\endgroup$
    – Silvinha
    Nov 11, 2020 at 19:47
  • $\begingroup$ OK -- what you have done is correct but difficult to follow due to inconsistent use of notation and too many details. I'll show you below a more clear and concise proof of the reverse implication. You understand the math -- just need improve on clarity. $\endgroup$
    – RRL
    Nov 11, 2020 at 19:50

1 Answer 1

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Suppose $B_1, \ldots, B_n$ are non-overlapping blocks (closed rectangles) such that $A = \cup_{k=1}^n B_k$ and $f|_{B_k}$ is Riemann integrable for all $k$. For any $\epsilon > 0$ there are partitions $P_{B_1}, \ldots, P_{B_n}$ such that for $1 \leqslant k \leqslant n$,

$$U(f, P_{B_k}) - L(f, P_{B_k}) < \frac{\epsilon}{n}$$

The partitions of the individual blocks taken together form a partition $P_A$ of $A$ where

$$U(f,P_A) = \sum_{k=1}^n U(f, P_{B_k}), \quad L(f,P_A) = \sum_{k=1}^n L(f, P_{B_k}),$$

and, thus,

$$U(f,P_A) - L(f,P_A) = \sum_{k=1}^n ( U(f, P_{B_k})- L(f, P_{B_k})) < n \cdot \frac{\epsilon}{n} = \epsilon$$

Therefore, $f$ is Riemann integrable on $A$ (by the Riemann criterion).

Let $P_A$ be an arbitrary partition of $A$. Using vertices of the blocks $B_k$ we can construct a refinement $P'_A \supset P_A$ such that every sub-block of $P'_A$ is contained in some block $B_k$ and we have

$$L(f,P_A) \leqslant L(f,P'_A) = \sum_{k=1}^n L(f, P_{B_k})\leqslant \sum_{k=1}^n U(f, P_{B_k})= U(f,P'_A) \leqslant U(f,P_A)$$

This implies that, for any partition $P_A$,

$$L(f,P_A) \leqslant \sum_{k=1}^n\int_{B_k} f|_{B_k}\leqslant U(f,P_A)$$

Since $f$ has been shown to be Riemann integrable, for any $\epsilon > 0$ there is a partition $P_A$ such that $U(f,P_A) - L(f,P_A) < \epsilon$ and $L(f,P_A) \leqslant \int_Af \leqslant U(f,P_A)$.

Therefore, for every $\epsilon > 0$,

$$\left|\int_Af - \sum_{k=1}^n \int_{B_k}f|_{B_k} \right| < \epsilon \implies \int_Af = \sum_{k=1}^n \int_{B_k} f|_{B_k}$$

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  • $\begingroup$ wow, You saved me today!! Thank you very much. $\endgroup$
    – Silvinha
    Nov 11, 2020 at 20:04
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    $\begingroup$ @Silvinha: You're welcome. $\endgroup$
    – RRL
    Nov 11, 2020 at 22:08

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