This is an exercise question from the book Measure,Integration and Real Analysis by Sheldon Axler.

Suppose $ b < c$ and $A\subset(b,c)$. Prove that A is Lebesgue measurable if and only if $ | A | + |( b, c ) \setminus A | = c − b.$

Here $|A|$ represents outermeasure of A.
I'm trying to prove the converse part.
Definition of Lebesgue Measurable set : A Set A is said to be Lebesgue Measurable if $ \forall \epsilon>0 , \exists F(closed set) \subset A$ such that $|A\setminus F|<\epsilon $ I'm trying to prove this by contradiction.
Suppose A isn't Lebesgue Measurable. $\exists \epsilon>0$ such that $|A\setminus F|>\epsilon $ for all closed $F\subset A$.
Similarly $(b,c)\setminus A$ isn't Lebesgue measurable $\exists \epsilon>0$ such that $|((b,c)\setminus A) \setminus F|>\epsilon $ for all closed $F\subset (b,c)\setminus A$
Since (b,c) is Lebesgue measurable, $ \forall \epsilon>0 , \exists F(closed set) \subset (b,c)$ such that $|(b,c)\setminus F|<\epsilon $
If F is closed subset of A and E is closed subset of (b,c)\A then $|(b,c)\setminus (F\cup E)| = |A\setminus F|+|((b,c)\setminus A)\setminus E|$

I dont know how to arrive at a contradiction from this.

  • $\begingroup$ @Zerox OP gave definition. why are you questioning it? it changes from book to book. $\endgroup$ Commented Dec 19, 2020 at 8:18

2 Answers 2


Welcome to MSE!

You mention the definition of measurability in terms of being "almost closed", however Axler actually gives a whole slew of equivalent definitions of measurable. These show up in the section 2D, and seeing as your exercise is 2D.12, it makes sense that you might want to use these definitions too! I've included a screenshot for reference:

equivalent definitions of measurability

A proof that these are equivalent can be found on page 53 of Axler's book. Again, this is all done before your exercise of interest, so I will freely use these equivalent definitions.

Let's show the easy direction first, you say you've done it, but I'll go ahead and do it too for completeness:

If $A$ is measurable, why is $|A| + |(b,c) \setminus A| = c-b$?

From chapter 2D again (Theorem 2.72), we know that $|\cdot|$ is a measure on the measurable sets (so at least the name makes sense :P). But from the definition of a measure (chapter 2C), we have additivity of the measures of disjoint sets.

In particular, since $A$ and $(b,c) \setminus A$ are disjoint, we have

$$|A| + |(b,c) \setminus A| = |A \cup ((b,c) \setminus A)| = |(b,c)| = c-b$$

What about the other direction? This is notably trickier, as I'm sure you're aware. To show it, we have to use a trick which will follow you throughout measure theory: go back to the basics. In measure theory even more than other areas of math, we define properties $P$ first for "simple" objects. Then we show how to approximate arbitrary objects by simple ones, and define $P$ for arbitrary objects to be the limit of $P$ of the simple objects. Oftentimes there's a follow-up step, where we restrict attention from all objects to only those objects on which $P$ is "nice".

We have already seen that for the definition of a measure.

  • First we defined $|\cdot|$ for open intervals.
  • Then we extended the definition to arbitrary sets (this is the "outer measure")
  • Then we restrict attention to those sets on which the outer measure behaves well (we call these sets "measurable")

You'll see soon enough that we do the same thing with integration!

Ok, but why bring this up? Because when we have these kinds of iterative definitions, it's often a good idea to go back to basics.

In particular, recall (definition 2.2)

$$|E| \triangleq \inf \left \{ \left . \sum_n |I_n| ~\right |~ E \subseteq \bigcup_n I_n \right \}$$

where the $I_n$ are open intervals, and we've already defined $|(a,b)| = \ell((a,b)) = b-a$.

But what does this buy us? For any set $E$ any $\epsilon$, we can find some intervals $I_n^\epsilon \subseteq (b,c)$ so that $E \subseteq \bigcup_n I_n^\epsilon$ and $\sum_n |I_n^\epsilon| \leq |E| + \epsilon$ (do you see why?).

Working with a sequence of intervals is kind of unwieldy, though. So we'll simplify with subadditivity. We'll suggestively write $G^\epsilon \triangleq \bigcup_n I_n^\epsilon$. Then $E \subseteq G^\epsilon$, and we know

$$|G^\epsilon| = \left | \bigcup_n I_n^\epsilon \right | \overset{\star}{\leq} \sum_n |I_n^\epsilon| \leq |E| + \epsilon$$

Here $\star$ follows from subadditivity of outer measure (Theorem 2.8).

Notice also that $G^\epsilon$ is a union of open intervals. In particular it's open.

Now we make one last step, and define $B = \bigcap_n G^{1/n}$. Then we have $E \subseteq B$ (do you see why?), with $B$ borel (do you see why?), and $|E| = |B|$. This last statement may be tricky as you're gaining experience with analysis, so I'll give a (do you see why?), but I'll also include a proof here:

$E \subseteq B$, so by theorem 2.5 we have $|E| \leq |B|$. In the other direction, notice for every $n$, $E \subseteq G^{1/n}$, so $$|B| \leq |G^{1/n}| \leq |E| + \frac{1}{n}$$ But that means $|B| \leq |E|$, as needed.

So, at the end of the day, what have we done? We've proven a lemma that says

For any set $E$, There is a borel set $B \supseteq E$ so that $|E| = |B|$.

We're in the home stretch now! Applying the lemma shows that we can find a set $B \supseteq A$ so that $|B| = |A|$. But by criterion (g) above, if we can show $|B \setminus A| = 0$, then $A$ is measurable and we're done! This doesn't seem that far off at all!

We're left with one last trick: We must also approximate $A^c$. That is, let's apply the lemma again to get a borel set $C \supseteq A^c$ with $|C| = |A^c|$.

Now we finally apply our assumption: $|A| + |A^c| = |(b,c)|$. This tells us that $|B| + |C| = |(b,c)|$ too.

But since $C$ is measurable (it's borel), we also know that $|C| + |C^c| = |(b,c)|$.

Thus $|C| + |C^c| = |(b,c)| = |B| + |C|$, and (since everything in sight is finite), we have $|C^c| = |B|$, and both of these sets are borel! Notice also that $C^c \subseteq A \subseteq B$.

Finally, then $|B \setminus A| \overset{\heartsuit}{\leq} |B \setminus C^c| \overset{\star}{=} |B| - |C^c| = 0$. And we're done by condition (g). And one last (do you see why $\heartsuit$ and $\star$ hold?) for the road.


Also, this looks rather long and unwieldy, but that's primarily because I'm trying to provide motivation for every step. Once you become fluent with these concepts, the proof is quite short! Alex Ravsky's answer does a great job showing this.

I hope this helps ^_^

  • $\begingroup$ Hi, sorry to dig up an old question but I am interested in the same problem that you have solved and I have a question about your (great!) answer: in it you say that given any set $A\subset\mathbb{R}$ one can find a Borel set $B$ such that $B\supset A$ and $|B|=|A|$: now, this is clear to me, but at a certain point you also say about $A\subset (b,c)$ that such a $B$ can be found that is a subset of $(b,c)$ so my question is: why is that? What stops every Borel set $B\subset (b,c)$ from having the property that $|B|>|A|$? Thanks $\endgroup$
    – lorenzo
    Commented Sep 10, 2021 at 18:47
  • $\begingroup$ You state this in your answer in this way: "For any set $E$ any $\epsilon$, we can find some intervals $I_n^\epsilon \subseteq (b,c)$ so that $E \subseteq \bigcup_n I_n^\epsilon$ and $\sum_n |I_n^\epsilon| \leq |E| + \epsilon$". Why should such intervals $I_n^\epsilon \subseteq (b,c)$ exist? $\endgroup$
    – lorenzo
    Commented Sep 10, 2021 at 18:53
  • 2
    $\begingroup$ @lorenzo -- Well, if you're happy that some such borel set $B$ exists, you just aren't convinced that we can keep it inside $(b,c)$, then we can look at the intersection $B \cap (b,c)$ (which is borel and contained in $(b,c)$. Since $A \subseteq B$ and $A \subseteq (b,c)$, we have $A \subseteq B \cap (b,c)$. Then $|A| \leq |B \cap (b,c)|$ is clear by monotonicity. For the other direction, we know (again by monotonicity) that $|B \cap (b,c)| \leq |B| = |A|$. So $B \cap (b,c)$ gets the job done. $\endgroup$ Commented Sep 10, 2021 at 20:27
  • 1
    $\begingroup$ now it is perfectly clear, thank you so much! $\endgroup$
    – lorenzo
    Commented Sep 10, 2021 at 21:16
  • 1
    $\begingroup$ @manifold -- exactly! The idea is that $x_* = \inf \{ x_\alpha\}$ exactly when $x_*$ is "smaller, but not much smaller" than all the $x_\alpha$s. Precisely, we know $x_* \leq x_\alpha$ always holds ("$x_*$ is smaller") and moreover for any $\epsilon$ we like, we can reverse this inequality for SOME $\alpha$: $x_\alpha \leq x_* + \epsilon$ ("$x_*$ is not much smaller"). Now applying this to the definition of $|E|$ as an infimum, we learn exactly that (for any $\epsilon$ we like) $\sum_n |I_n| \leq |E| + \epsilon$ for some $E \subseteq \bigcup_n I_n$. $\endgroup$ Commented Mar 29, 2023 at 5:15

Suppose that $ | A | + |( b, c ) \setminus A | = c − b$ and $\epsilon>0$ be any number. There exist countable covers $\mathcal C_1$ of $A$ and $\mathcal C_2$ of $( b, c ) \setminus A$ by open subintervals of $(b,c)$ such that $S_1+S_2\le c-b+\epsilon$, where $S_i=\sum_{C\in\mathcal C_i} |C| $ for each $i$. It is easy to check that if $|\bigcup\mathcal C_1\cap\bigcup \mathcal C_2|>\epsilon$ then $\mathcal C_1\cap\mathcal C_2$ cannot cover $(b,c)$, so $|\bigcup\mathcal C_1\cap\bigcup \mathcal C_2|\le\epsilon$. Put $F=(b,c)\setminus\bigcup \mathcal C_2$. Clearty $F$ is a closed set contained in $A$. Then $A\setminus F= A\cap \bigcup \mathcal C_2\subset \bigcup \mathcal C_1\cap\bigcup \mathcal C_2$, so $| A\setminus F |\le\epsilon$.


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