# Integrability of the function on the intersection of the Lebesgue measure $0$

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.

• Which claim are you refering to in your second question? Mar 31, 2022 at 21:13
• @blamethelag, I have only one question, if the assumption is redundant in the theorem, or $f$ can fail to be integrable on $C\cap E$ Apr 1, 2022 at 4:26
• How do you define Riemann integrability on Lebesgue measurable sets? Apr 1, 2022 at 14:31
• @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 $\varepsilon$. Apr 1, 2022 at 15:10
• 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. Apr 1, 2022 at 17:35

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 $$C \cap E$$.

• 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 $$\max\{f(x),g(x)\}=\frac{f(x)+g(x)}2+\frac{|f(x)-g(x)|}2\\\min\{f(x),g(x)\}=\frac{f(x)+g(x)}2-\frac{|f(x)-g(x)|}2$$ Apr 8, 2022 at 6:15
• @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 $x \in C^C \cap E^C$.
– RRL
Apr 8, 2022 at 16:08