# On generical domains: Riemann integrable $\Rightarrow$ Lebesgue integrable?

If I have correctly understood and been able to generalise the proof that I found in Kolmogorov-Fomin's Элементы теории функций и функционального анализа for the case of $n=1$, I know that, if $R=\prod_{i=1}^n[a_i,b_i]\subset\mathbb{R}^n$ and $f:R\to\mathbb{R}$ is Riemann integrable on the bounded set $R$, then it also is Lebesgue integrable and the two integrals are the same:$$(R)\int_R f(x_1,\ldots,x_n)dx_1\ldots dx_n=\int_R fd\mu$$where $\mu$ is the usual Lebesgue measure defined on measurable subsets of $\mathbb{R}^n$.

$f:D\to\mathbb{R}$ is defined as Riemann integrable on a generical domain $D\subset\mathbb{R}^n$ when the function $\bar{f}:R\to\mathbb{R}$ (with $R$ as above and such that $D\subset R$) defined by $$\bar{f}(\boldsymbol{x}) = \begin{cases} f(\boldsymbol{x}), & \boldsymbol{x}\in D \\ 0, &\boldsymbol{x}\in R\setminus D \end{cases}$$is Riemann integrable on $R$. My question is: if $f$ is Riemann integrable on a generical domain $D$, can we say that it also is Lebesgue integrable on $D$? I think the only obstacle to overcome is veryfying whether $D$ is $\mu$-measurable, because Lebesgue integrals are only defined on measurable subsets, as far as I know, but I cannot overcome that. A search in this site gives many related results, but I have only found cases where $D=[a,b]\in\mathbb{R}$. I thank any answerer very much!

• What do you mean by "generical domain"? The term is not in the Wikipedia page you point to. Refer also to Baby Rudin 11.33 showing that in one dimension, a Riemann integrable function as he defines it is one which is Lebesgue integrable and continuous almost everywhere. Note too that there is some wobble in the definition of Riemann integrable functions; some say that Riemann integrable functions must be bounded and defined on bounded domains, others that "Riemann integrable" includes extensions of the Riemann integral by taking limits (e.g., for $\int_0^\infty f$ take lim $\int_0^a f$). – ForgotALot Jan 17 '16 at 22:21
• On further reflection it seems that what you are asking is if $D\subset R$ is nonmeasurable, and $\overline f=1_D f$ is nonetheless Riemann integrable, is it also Lebesgue integrable? In one dimension, the reasoning in Baby Rudin 11.33 seemingly applies and shows that yes, $\overline f$ is Lebesgue integrable. You could have $D$ nonmeasurable yet $1_D f$ Riemann integrable if, for example, $f:[a,b]\to\mathbb{R}$, $f(x)=0$ for $x\in [a,(a+b)/2]$, and $D\subset [a,(a+b)/2]$. – ForgotALot Jan 17 '16 at 22:48
• Thank you for your comments. What I call "generical domain" $D\subset R$ (with $R$ bounded and defined as above) is what the Wikipedia page I have linked calls "an arbitrary bounded n-dimensional set". I did not know that if $D$ is not $\mu$-measurable we can nevertheless define a Lebesgue integral $\int_Dfd\mu$, since Kolmogorov-Fomin's defines it only for $\mu$-measurable domains. – Self-teaching worker Jan 18 '16 at 0:34
• According to this definition of Lebesgue integral, I would think (taking into account that $\bar{f}$ is $\mu$-measurable if it is Rieman integrable on $R$) that if the Riemann integrability of $f$ on $D$ implied the $\mu$-measurability of $D$, then the Lebesgue integrability of $f$ would follow from $D$'s measurability; but you suggest that $D$ may not be measurable even if $f$ is Riemann integrable on it... – Self-teaching worker Jan 18 '16 at 0:34

It is a matter of definition.

Let's decide, as for Riemann, that $f$ is called $\,$ L-integrable over $D\,$ if $\,\bar f \,$ is $\,$L-integrable over $R$ (according to the usual definition) and that $$\int_D f=\int_{R} \bar f$$ Then the statement (R-integrability $\Rightarrow$ L-integrability) is true also for arbitrary (bounded) sets.

The set $D$ doesn't need to be measurable !

All this because of the validity of the following statement:

Let $E$ and $E'$ be measurable sets with $E \subset E'$. Let $f$ be a function on $E$ and let $g$ be the function defined on $E'$ by $$g(\mathbf x) = \begin{cases} f(\mathbf x)\quad\text {if} \quad \mathbf x\in E \\ 0 \;\,\qquad\text {if}\quad\mathbf x\in E'\setminus E \end{cases}$$ then $f$ is L-integrable on $E\,$ iff $\,g$ is L-integrable on $E'$ and, if so, $$\int_E f=\int_{E'} g$$ So the above definition is independent of $R$ and reduces to the usual one if $D$ is measurable.