QUESTION a) Let $f:A\rightarrow \mathbb{R}$ be a integrable function defined in a block $A\subset\mathbb{R}^m$. If $A_0\subset\mathbb{R}^m$ is a block such that $A\subset A_0$, then the function $\overline{f}:A_0\rightarrow \mathbb{R}$ given by

\begin{equation*} \overline{f}(x) = \begin{cases} f(x) & \text{if } x \in A,\\ 0 & \text{if } x \in A_0\backslash A. \end{cases} \end{equation*} is integrable (in $A_0$) and $$\int_{A_0}\overline{f}(x) dx= \int_{A} f(x) dx$$

QUESTION b) Let $A=\displaystyle\Pi_{i=1}^{n}[a_i, b_i]$ be a block in $\mathbb{R}^m$, such that $A \subset A_0$, then $$\int_{A_0}\chi_{{}_{A}} dx = \Pi_{i=1}^{m} (b_i-a_i)$$


a) Because $A\subset A_0$ let's divide in two cases:

  1. Supposing that $A_0\cap A=A$ by definition of $f$ we have $$\int_{A_0} \overline{f}(x) dx = \int_{A} f(x) dx.$$

  2. Now, in the set $A_0\backslash A$ we have $\overline{f}(x) = f(x) =0$. Considering $A_0\backslash A \subset \displaystyle\bigcup_{i=1}^{\infty} C_i$, where $C_i's$ are open blocks, by the Lindelöf's theorem has a countable subcover, let's define it as $A_0\backslash A\subset \displaystyle\bigcup_{i=1}^{k} A_i =D$, however all countable set has zero measure (because is a set of points which has zero measure). Then we can write $\{x \in A_0\backslash A; f(x)=0\}= D$. Thus $$\int_{A_0}\overline{f}(x) dx = \int_{A} f(x) dx.$$

b) Supposing $A\subset A_0$, let's divide in two cases:

  1. if $\overline{A_0}=\partial A$, so because $\partial A$ is defined as an edge of the block $A=\Pi_{i=1}^{n}[a_i, b_i]\subset \mathbb{R}^m$, then by definition has zero measure. Thus in this case, $$\int_{\overline{A_0}} \chi_{{}_{A}}(x) dx = \int_A \chi_{{}_{A}}(x) dx= \Pi_{i=1}^{m}(b_i-a_i).$$

  2. if $A$ is a proper subset of $A_0$, then we have by definition of characteristic function that $\chi_{{}_{A_{0}\backslash A}}(x) = 0$. Therefore, by the question one, we have $$\int_{A_0} \chi{{}_{A}} dx = \int_{A} \chi{{}_{A}} dx = \Pi_{i=1}^{m}(b_i-a_i).$$

MY DOUBTS: Are my answers right? Would you correct for me or help me to improve it, please?


1 Answer 1


Your proof is fine, but it might be simpler just to observe that when you take upper and lower sums for the Jordan measure of a set, it's the same as taking upper and lower Darboux sums for a Riemann integral, so your integrals are simply Riemann integrals over cubes. And since it's given that $f$ is integrable, and $\chi_{A}$ certainly is, then $g:=\overline f\cdot \chi_{A}$ is also integrable, and this gives $a).$ For $b).\ $ we go the other way, that is, integrals of the constant function $f=1$ over cubes give Jordan content, so we have $\int_{A_0} \chi{{}_{A}} dx=\int_{A} dx=\operatorname{vol} A=\Pi_{i=1}^{m}(b_i-a_i).$


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