4
$\begingroup$

Suppose $C,E\subset\Bbb R^2$ are bounded s. t. $C\cap E$ is of Lebesgue measure $0$ and let $A\supseteq C\cup E.$ If $f:A\to\Bbb R$ is (Riemann) integrable on $C$ and $E,$ is it necessarily (Riemann) integrable on $C\cap E$?

This question arose when I was revising the theorem about the additivity of an integral w.r.t. to the area of integration:

$\underline{\boldsymbol{\text{ theorem }8.9.}:}$

Suppose $C,E\subset\Bbb R^2$ are bounded s. t. $C\cap E$ is of Lebesgue measure $0$ and that $f:C\cup E\to\Bbb R$ is a function integrable on $C,E,$ and $C\cap E$. Then $f$ is integrable on $C\cup E$ and $$\int_{C\cup E}f=\int_C+\int_Ef.$$

Just in case, I'm going to write down the proof for the context:

  Let $A$ be a rectangle containing $C\cup E$ and let's note the extension of $f$ by $0$ from $C\cup E$ to $A$ by $\tilde f.$ We're going to observe the characteristic functions $\chi_C,\chi_E$ and $\chi_{C\cap E}$ of the sets $C,E$ and $C\cap E$ as functions on $A$. Let $f_1=\tilde f\chi_C,f_2=\tilde f\chi_E$ and $f_3=\tilde f\chi_{C\cap E}.$ Then  $f_1,f_2$ and $f_3$ are extensions by $0$ of  $f_{\mid  C},f_{\mid E}$ and $f_{\mid C\cap E}$ on $A,$ respectively. Therefore, $f_1,f_2$ and $f_3$ are integrable on $A.$ Clearly $\chi_{C\cup E}=\chi_C+\chi_E-\chi_{C\cap E}.$ It follows $$\tilde f=\tilde f\chi_{C\cup E}=f_1+f_2-f_3.$$ Hence, $\tilde f$ is integrable on $A$ and it holds $\int_A\tilde f=\int_Af_1+\int_Af_2-\int_Af_3.$ Since $\int_A\tilde f=\int_{C\cup E},\int_Af_1=\int_Cf,\int_Af_2=\int_Ef$ and $\int_Af_3=0,(*)$ the claim follows.

$(*)$ is the result proven shortly before:

$\underline{\boldsymbol{\text{ proposition }8.6}}:$

If $C\subset\Bbb R^2$ is bounded and of Lebesgue measure $0$ and $f:C\to\Bbb R$ is integrable on $C,$ (hence, bounded) then $\int_Cf=0.$

Question:

Would the same claim hold without the assumption that $f$ is (Riemann) integrable on $C\cap E,$ that is, is that assumption redundant if other assumptions of the theorem remain?

I thought of first partitioning $C$ and $E$ separately and then $C\cap E$ as the subset of which and then find the common refinement of $P_C$ and $P_{C\cap E}$ and of $P_E$ and $P_{C\cap E}$ and comparing Darboux sums, however I haven't come up with anything. An idea of taking some rectangle $B$ containing $C\cap E$ and an arbitrarily small stripe around $\partial B$ was also floating around...


EDIT (for clarification)

That $f$ is (Riemann) integrable on $C\cap E$ is part of the assumption of the above stated theorem. I was wondering whether that assumption was redundant if $f$ might necessarily be (Riemann) integrable on $C\cap E,$ so my question boils down to:

If a function $f$ is (Riemann) integrable on two bounded sets $C,E\subset\Bbb R^2,$ and their intersection is of Lebesgue measure $0,$ is it necessarily (Riemann) integrable on the intersection, too?

Now that I wrote this, if we forget the assumption on the intersection being of Lebesgue measure $0,$ would that affect integrability?

EDIT ( $6^{\mathrm{th}}$ April 2022):

I would like to include the lemma:

$\underline{\boldsymbol{\text{ lemma }8.8:}}$

Suppose $C\subseteq\Bbb R^2$ is a set of Jordan measure $0$ and $f: C\to\Bbb R$ is bounded. Then $f$ is integrable and $\int_Cf=0.$

Maybe now my question can boil down to:

If $C,E\subset\Bbb R^2$ are bounded, $C\cap E$ is of Lebesgue measure $0,$ and $f:C\cup E\to\Bbb R$ is integrable on both $C$ and $E,$ is it possible that $C\cap E$ has no Jordan measure, that is, $\chi_{C\cap E}$ isn't integrable on any rectangle $A\supset C\cap E$?

Because if $C\cap E$ is of Lebesgue measure $0$ and has Jordan measure, its Jordan measure is necessarily $0$ and $f$ is integrable on $C\cap E$ by the lemma 8.8.

$\endgroup$
12
  • $\begingroup$ Which claim are you refering to in your second question? $\endgroup$ Mar 31, 2022 at 21:13
  • $\begingroup$ @blamethelag, I have only one question, if the assumption is redundant in the theorem, or $f$ can fail to be integrable on $C\cap E$ $\endgroup$
    – PinkyWay
    Apr 1, 2022 at 4:26
  • $\begingroup$ How do you define Riemann integrability on Lebesgue measurable sets? $\endgroup$
    – Jakobian
    Apr 1, 2022 at 14:31
  • $\begingroup$ @Jacobian, we defined only sets of Lebesgue measure $0$. We said a bounded function $f:A\to\Bbb R$ from a bounded set $A\subset\Bbb R^2$ is Riemann integrable its extension $\tilde f$ by $0$ is Riemann integrable on any rectangle $B\supseteq\Bbb R^2$ and $$\int_Af=\int_B\tilde f.$$ We said a set $S\subseteq\Bbb R^2$ is of (Lebesgue) measure $0$ if for any $\varepsilon>0, S$ can be covered with at most countably many rectangles of the overall area less than $\varepsilon$. $\endgroup$
    – PinkyWay
    Apr 1, 2022 at 15:10
  • $\begingroup$ A bounded function $f:[a,b] \longrightarrow \mathbb R$ is Riemann integrable if and only if it is Lebesgue integrable and is Lebesgue almost everywhere continuous. If you can replace $[a,b]$ by a rectangle you are done. $\endgroup$ Apr 1, 2022 at 17:35

1 Answer 1

3
+100
$\begingroup$

We have the following proposition (regardless of the measure of $C \cap E$).

If $f$ is Riemann integrable on bounded sets $C,E \subset \mathbb{R}^2$, then $f$ is Riemann integrable on $C \cap E$.

The proof will rely on the following lemma (which you should be able to prove in a straightforward way by showing that the min and max preserve continuity.)

Let $g(x) = \max(f_1(x), f_2(x)$) and $h(x) = \min(f_1(x), f_2(x)$) where $f_1$ and $f_2$ are Riemann integrable on a set $S\subset \mathbb{R}^2$. Then $g$ and $h$ are Riemann integrable on $S$.

Proof of proposition.

Let $f^+(x) = \max (f(x),0)$ and $f^-(x) = \max (-f(x),0)$. Then $f^+$ and $f^-$ are nonnegative functions such that $f(x) = f^+(x) - f^-(x)$. If $f$ is Riemann integrable on $C$ and $E$, then it follows from the lemma that $f^+$ and $f^-$ are Riemann integrable on $C$ and $E$.

Using the notation $f^+_S := f^+\chi_S$ we have, since $f^+$ is nonnegative,

$$f^+_{C\cap E}(x) = \min(f^+_C(x), f^+_E(x))$$

SInce $f^+$ is Riemann integrable on $C$ and $E$, we have $f^+_C$ and $f^+_E$ Riemann integrable on a rectangle $A$ containing $C$ and $E$ (and, hence, $C\cap E$). By the lemma it follows that $f^+_{C\cap E}$ is Riemann integrable on A and, by definition, $f^+$is Riemann integrable on $C \cap E$.

By the same argument we can show that $f^-$ is Riemann integrable on $C \cap E$ and by integral additivity $f = f^+-f^-$ is Riemann integrable on $C \cap E$.

$\endgroup$
2
  • $\begingroup$ Thank you very much! This bothered me for a week. May I ask, did you use the fact $\chi_{C\cap E}=\min\{\chi_C,\chi_E\}$? I expanded $$\begin{aligned}f^+_{C\cap E}(x)&=\max\{f(x),0\}\cdot \chi_{C\cap E}(x)\\&=\max\{f(x),0\}\cdot\min\{\chi_C,\chi_E\}\\&=\min\{\max\{f(x),0\}\chi_C,\max\{f(x),0\}\chi_E\}\\&=\min\{f^+_C,f^+_E\}\end{aligned}$$ and for the first part, I used $$\max\{f(x),g(x)\}=\frac{f(x)+g(x)}2+\frac{|f(x)-g(x)|}2\\\min\{f(x),g(x)\}=\frac{f(x)+g(x)}2-\frac{|f(x)-g(x)|}2$$ $\endgroup$
    – PinkyWay
    Apr 8, 2022 at 6:15
  • 1
    $\begingroup$ @Spring: You're welcome. Thanks for filling in that detail. I simply observed that $f^+_{C\cap E} = \min(f^+_C, f^+_E)$ by looking at the mutually exclusive and exhaustive cases $x \in C\cap E$, $x \in C \cap E^C$, $x \in C^C\cap E$, and $x \in C^C \cap E^C$. $\endgroup$
    – RRL
    Apr 8, 2022 at 16:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .