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

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.

• 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.
– RRL
Nov 11, 2020 at 19:24
• Yes, you're write isn't lebesgue-integral. Nov 11, 2020 at 19:47
• 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.
– RRL
Nov 11, 2020 at 19:50

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 $$\epsilon > 0$$,

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

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