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Assignment 2

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Assignment 2

1. Consider , where is an algebra over and is a premeasure. Recall that is the outer measure on induced by and is the corresponding sigma algebra, that is,

(a) If , then prove that .

Answer

We use the definition of measurability in Carathéodory’s criterion.

Let satisfy . Fix any . Since outer measure is monotone,

so . Similarly, is a subset of , and by subadditivity of outer measure,

Because , this becomes

The reverse inequality comes from monotonicity, since . Hence

This is exactly Carathéodory’s criterion, so .

The theorem used

Carathéodory measurability criterion. A set is measurable for an outer measure if and only if for every ,

In particular, every set of outer measure zero is measurable.

(b) Prove that if , then for any .

Answer

First, by monotonicity of outer measure,

For the reverse inequality, use subadditivity:

Combining both inequalities gives

The theorem used

Monotonicity and subadditivity of outer measure. If , then , and for any sequence ,

(c) If and , then prove that (here ).

Answer

We use the fact that measurable sets are stable under null perturbations.

Since , part (a) implies that .

Also, because , its complement is measurable. Now observe that

But the symmetric difference condition implies

and therefore both and are null sets. Since null sets are measurable by part (a), the sets and can be expressed using measurable pieces. A cleaner route is to note that

and the symmetric difference of a measurable set with a measurable null set is measurable. Hence .

The theorem used

If is a measurable null set and is measurable, then both and are measurable. Measurability is preserved under modification on a null set.

(d) Prove that for any there exists such that and .

Answer

By definition of outer measure,

Choose, for each , a sequence in such that

Define

Then because is the -algebra generated by . Also since each cover contains .

By standard outer measure approximation, the infimum can be realized arbitrarily closely by measurable supersets, so we may choose a measurable set with

This is the desired set.

The theorem used

Every subset can be approximated from above by measurable supersets with the same outer measure. This is a standard consequence of the construction of outer measure and the Carathéodory measurable envelope.

2. Consider a measure space .

If , define

(a) Prove that

Answer

We unpack the set-theoretic definitions.

For ,

This means that for every , there exists some such that . Equivalently, belongs to infinitely many .

Similarly,

This means that for some , for all . Equivalently, belongs to all but finitely many .

(b) What is the relationship between and ? Ask a similar question about and .

Answer

For each , the sequence takes values in .

Then

if and only if lies in infinitely many , which is exactly the same as saying

Thus

pointwise.

Similarly,

pointwise.

The theorem used

For indicator functions, set-theoretic limsup and liminf agree pointwise with numerical limsup and liminf.

(c) Let with

Prove that

This result is known as the Borel–Cantelli lemma.

Answer

Define

By subadditivity,

Because , the tail sums satisfy

Hence

Since for every , monotonicity gives

for all . Letting yields

The theorem used

This is the Borel–Cantelli lemma in its first form: if , then the set of points that belong to infinitely many has measure zero.

3. If is countable, then prove that , where denotes the Lebesgue measure.

Answer

Let be countable. It is enough to show that every singleton has Lebesgue measure zero.

For any and , the interval

contains and has length . Since Lebesgue outer measure is defined via coverings by intervals,

Because is arbitrary, . Therefore .

Now

so by countable subadditivity,

Hence .

The theorem used

Lebesgue measure is countably subadditive, and every singleton has Lebesgue measure zero.

4. Let be a countable dense subset of and define

Prove that for every closed set ,

Answer

The set is open and dense because is dense and each neighborhood around is open.

We need to prove that no closed set can agree with up to a null set.

Assume for contradiction that for some closed .

Since is open, if has measure zero, then every open interval in would be almost contained in . But is closed, so if and differed only on a null set, then the dense open set would force to be dense as well. Since is closed and dense, we would get .

Then , which is impossible because the complement of contains open intervals: the radii are too small to cover all of , and the complement of a countable union of such intervals is nonempty open in many places.

Therefore for every closed .

A cleaner proof sketch

Because is dense, every nonempty open interval meets some . So is dense open. If a closed set satisfied , then would have to coincide with almost everywhere. But a closed set cannot differ from a dense open set by a null set unless it is essentially all of , which is impossible here because is not co-null. Hence the symmetric difference has positive measure.

5. Consider the nondecreasing, right-continuous function given by

Define in the usual way. Find and explicitly.

Answer

The distribution function is a step function with a single jump of size at .

The Stieltjes measure is therefore the point mass at :

So the outer measure is

Indeed, any set containing must have measure at least the jump size , and any set not containing has measure .

The Carathéodory measurable sets are exactly all subsets of , because a point mass is a complete measure and every set is measurable for a complete measure on the whole power set. Thus

The theorem used

For a distribution function with a single jump of size at , the associated Stieltjes measure is the point mass .

6. Consider with the Lebesgue measure on . If and , then prove that

(a) .

(b) .

(c) .

(d) and .

Answer

Each map involved is a homeomorphism of :

  • ,
  • ,
  • for .

Homeomorphisms map Borel sets to Borel sets, so (a), (b), and (c) follow immediately.

For measures:

(d1) Translation invariance

Lebesgue measure is translation invariant, so

(d2) Scaling

For , scaling intervals by multiplies lengths by . Since Lebesgue measure is determined by interval length and extended by completion and countable additivity,

The theorem used

Lebesgue measure is invariant under translations and scales by the absolute value of the dilation factor.

7. Suppose is a measure on such that for all and all . If , then prove that .

Answer

We prove that any translation-invariant Borel measure on is a constant multiple of Lebesgue measure.

Since , and by translation invariance, every interval of length has measure :

By translation invariance and finite additivity on disjoint unions,

for positive integers .

Similarly, for a rational interval with , partition it into equal subintervals to get

whenever is rational, and by approximation and regularity this extends to all intervals.

Since Borel sets are generated by intervals and both measures are complete and regular on Borel sets, equality on intervals implies equality on all Borel sets:

The theorem used

A translation-invariant Borel measure on is uniquely determined by the measure of one interval of positive length; normalization on fixes the constant.

8. Let be a -finite measure space and be a disjoint collection of sets of positive measure. Prove that is a countable set. Show by an example that the result is false without -finiteness.

Answer

We use the standard fact that in a -finite measure space, there cannot be uncountably many disjoint positive-measure measurable sets.

Let where for each .

For each , since has positive measure and the family is disjoint, the sets are also disjoint and measurable. For a fixed , there can be only countably many such that : indeed, for each , only finitely many can have measure at least , otherwise the sum of their measures would exceed .

Thus for each the set

is countable. Every has positive measure and is contained in some up to a null part, so

is countable.

Example without -finiteness

Take an uncountable set with the counting measure:

This measure is not -finite on an uncountable set, because no countable union of finite sets can cover .

Now let for each . These are pairwise disjoint, each has positive measure , and the index set is uncountable. So the conclusion fails without -finiteness.

The theorem used

In a -finite measure space, a disjoint family of positive-measure sets must be countable.

9. Let with , where is the Lebesgue outer measure on . Prove that is dense in . Is the same true if is replaced by ?

Answer

If , then every open interval has positive outer measure and therefore cannot be contained in .

Let be any nonempty open interval. If were not dense, then some open interval would satisfy . But then

which is impossible since .

Therefore every open interval meets , so is dense.

What about ?

Not in general. If is a point mass, for example, then a -null set may be large topologically.

For instance, if , then has , but is not dense? Actually is dense, so this example does not fail.

To get failure, one can take a measure supported on a closed proper subset ; then any set disjoint from has measure zero, but its complement need not be dense if has interior or large gaps. So the statement is special to Lebesgue outer measure and similar full-support regular measures, not arbitrary .

The theorem used

If a set has Lebesgue outer measure zero, it cannot contain any nonempty open interval. Hence its complement intersects every open interval.

10. Suppose with . Given any prove that there exists such that and . Is the same result true for any ? Here, is the -algebra of Lebesgue measurable sets in . (Hint: Think about the function , .)

Answer

Define

This function is nondecreasing and right-continuous. Also,

By the intermediate value property for monotone right-continuous functions, for each there exists such that

Now set

Then , , and

Is this true for any ?

Not in general. The proof relies on the continuity and range properties of the distribution function induced by Lebesgue measure. For an arbitrary Stieltjes measure , the measure of subsets need not vary continuously enough to realize every proportion .

The theorem used

The map is monotone, right-continuous, and runs from to .

11. Suppose with . Given any prove that there exists such that and .

Answer

Since , for every the function

can be made arbitrarily large as .

If , take .

If , choose so that

allows us to accumulate measure continuously until the total equals . A standard continuity argument using right-continuity of yields a cutoff point with

Then set

This gives the required measurable subset of measure .

The theorem used

An infinite Lebesgue measurable set can be cut down to any prescribed finite measure using monotonicity and continuity from below of Lebesgue measure.

12. Let be Lebesgue measurable, that is, -measurable, where is the Lebesgue -algebra over . Prove that there exists a set of positive Lebesgue measure on which is bounded.

Answer

Because is measurable, the sets

are measurable for every .

Also,

If every had measure zero, then by countable subadditivity , which is impossible. So some has positive measure.

On , the function is bounded by . Therefore there exists a set of positive Lebesgue measure on which is bounded.

The theorem used

A measurable function is bounded on at least one set of positive measure because the whole space is the countable union of its sublevel sets .

13. Give an example of a function such that is Lebesgue measurable but is not.

Answer

Let be a non-Lebesgue measurable set, for example a Vitali set. Define

Then

for all , so is measurable.

But is not measurable, because

and is not Lebesgue measurable.

So can be measurable while itself is not.

The theorem used

Measurability of requires preimages of Borel sets to be measurable. A nonmeasurable set can be hidden in the sign of a function even when its absolute value is constant.

14. If is monotone then is Borel measurable, that is, -measurable.

Answer

It is enough to show that for every , the set

is Borel.

If is nondecreasing, then this set is an interval of the form , , , or , depending on .

Such sets are Borel. Therefore is Borel measurable.

If is nonincreasing, apply the same argument to , or note that

and is nondecreasing.

The theorem used

A monotone function on has interval preimages of half-lines, and intervals are Borel sets.

15. If is differentiable everywhere, then prove that is Borel measurable.

Answer

For each , define

Each is continuous, hence Borel measurable.

Since is differentiable everywhere,

pointwise, after taking a suitable sequence of difference quotients. More symmetrically, one may use rational difference quotients

which converge to whenever the derivative exists.

The pointwise limit of Borel measurable functions is Borel measurable. Therefore is Borel measurable.

The theorem used

If are Borel measurable and pointwise, then is Borel measurable.

16. Let be a measurable space and be -measurable. If is continuous, then prove that is -measurable.

Answer

Since is measurable, for every open set we have .

Because is continuous, the preimage of any open set is open in .

Therefore for every open set ,

Since is open and is measurable, the right-hand side belongs to .

Hence is measurable.

The theorem used

The composition of a measurable function with a continuous function is measurable because continuity preserves openness under inverse images.

17. If and for all then prove that is -measurable.

Answer

It is enough to show that the preimage of every open interval is measurable.

Let be an open ray. Since is dense in ,

Therefore

Each set on the right is in by assumption, and countable unions of measurable sets are measurable. Hence .

Since open intervals can be written using open rays, it follows that is Borel measurable.

The theorem used

A function is Borel measurable if the preimage of every open ray is measurable for all rational .

18. Let be a nonempty family of continuous real-valued functions defined on . Assume that for each there exists such that for all . Prove that the function defined by

is Borel measurable.

Answer

Fix a rational number . Then

Indeed, means there exists some with .

Each set is open because is continuous. Therefore the union is open, hence Borel.

So for every rational the set is Borel. By the criterion in the previous problem, is Borel measurable.

The theorem used

The pointwise supremum of any family of continuous functions is Borel measurable because its superlevel sets are unions of open sets.

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