Assignment 1

Assignment 1

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Question 1. Given a nonempty set , describe the smallest and largest -algebra of subsets of .

Answer

The smallest -algebra on a nonempty set is

The largest -algebra on is the power set

Why this is true

A -algebra on must satisfy:

1. .
2. If , then .
3. If , then .

Since , by closure under complements we also get . So every -algebra contains at least these two sets. The collection is itself a -algebra, so it is the smallest one.

On the other hand, contains every subset of , so it is automatically closed under complements and arbitrary unions. Hence it is a -algebra, and clearly the largest possible one.

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Question 2. Given , describe and , the algebra and the -algebra generated by the set .

Answer

The algebra generated by the single set is

The -algebra generated by is the same:

Explanation

To form the smallest algebra containing , we must include:

• itself,
• its complement ,
• the empty set ,
• and the whole space .

These four sets are closed under complements and finite unions, so they form an algebra. They are also closed under countable unions, because any countable union of sets from this family is again one of these four sets. Therefore the generated algebra and generated -algebra coincide.

Special cases

If or , then the family collapses to the smallest -algebra:

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Question 3. Let be a nonempty set. Define

Assume that is a finite set. Prove that is an algebra. But is not a -algebra if is an infinite set.

Answer

Step 1: is an algebra

We check the algebra axioms.

(i)

The empty set is finite, and is finite, so . Thus both and belong to .

(ii) Closed under complements

If is finite, then is cofinite, so .

If is finite, then is cofinite, hence is finite, so again .

So is closed under complements.

(iii) Closed under finite unions

Let . We consider cases.

• If both are finite, then is finite.

• If one is cofinite, say is finite, then

  so is finite, meaning is cofinite.

• If both are cofinite, then their union is cofinite as well.

Hence is closed under finite unions.

Therefore is an algebra.

Step 2: is not a -algebra when is infinite

Assume is infinite. Choose a countably infinite subset consisting of distinct points. Each singleton is finite, hence belongs to .

Now consider the countable union

This set is infinite. If were finite, then would be cofinite, which would force itself to be countable and would still not make finite. More importantly, since is infinite, we can choose the sequence so that is a proper infinite subset of . Then is neither finite nor cofinite. Therefore .

So is not closed under countable unions, hence it is not a -algebra.

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Question 4. Let be a nonempty set. Define

Prove that is a -algebra.

Answer

We verify the -algebra axioms.

(i) and belong to

• is countable.
• is countable, so is in .

(ii) Closed under complements

If is countable, then is co-countable, so .

If is countable, then is co-countable, so again .

So is closed under complements.

(iii) Closed under countable unions

Let .

There are two cases.

Case 1: Each is countable

Then

is a countable union of countable sets, hence countable. So the union belongs to .

Case 2: Some is co-countable

Then is countable. Since

the complement of the union is a subset of a countable set, hence countable. Therefore the union is co-countable, and so it belongs to .

Thus is closed under countable unions.

Therefore is a -algebra.

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Question 5. Let be a nonempty set and let be a partition of , that is, for all and . Let

where for , . Prove that is a -algebra.

Answer

We show that satisfies the three -algebra properties.

(i) and are in

• If , then .
• If , then .

So .

(ii) Closed under complements

Take any . Then for some .

Because the form a partition of ,

Thus .

(iii) Closed under countable unions

Let for each . Write

Then

Since , the right-hand side is again in .

Hence is a -algebra.

Conceptual meaning

This -algebra consists exactly of all unions of partition blocks. So the partition decides which blocks are kept and which are discarded.

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Question 6. Let be a nonempty set and be a -algebra on . For , define

Prove that is a -algebra on . This is called the trace -algebra of on .

Answer

We verify the -algebra axioms relative to the space .

(i) The whole set and the empty set are in

• Since , we have .
• Since , we have .

(ii) Closed under complements in

Take . Then for some .

The complement of inside is

Since , we have .

(iii) Closed under countable unions

Let for each , and write with .

Then

Since is a -algebra, . Therefore the union belongs to .

Hence is a -algebra on .

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Question 7. Consider the semi-algebra of semiclosed intervals in (see Lecture 2). Prove that , the Borel -algebra.

Answer

We prove two inclusions.

First inclusion:

Every semiclosed interval is a Borel set. For example, if consists of intervals of the form , then

and both and are Borel sets. So each element of is Borel, and therefore the -algebra generated by is contained in .

Second inclusion:

It is enough to show that every open interval is in , because open sets are countable unions of open intervals.

If , then

where empty sets may appear for large if .

Each lies in , so .

Since every open set is a countable union of open intervals, every open set belongs to . Hence .

Therefore

The key theorem used

The Borel -algebra on is the smallest -algebra containing all open sets. Since open intervals generate all open sets, any -algebra containing all open intervals must contain .

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Question 8. For above, how do we know that is not a -algebra?

Answer

The algebra generated by semiclosed intervals consists of finite unions of sets from . In particular, every set in has only finitely many interval components.

To show it is not a -algebra, we exhibit a countable union of members of that is not in .

Consider the family of intervals

Each one belongs to , so each is in the algebra. But their countable union is

and this set cannot be written as a finite union of semiclosed intervals from in the way elements of the algebra are built.

So is not closed under countable unions, and therefore it is not a -algebra.

Short conceptual reason

An algebra only guarantees closure under finite unions. A -algebra requires closure under countable unions, and finite unions are not enough to capture that extra closure.

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Question 9. Prove that every countable subset of is in .

Answer

Let be countable. Then we can write

or as a finite set if the sequence stops.

Each singleton is closed in , hence Borel. Indeed,

and both sets on the right are closed, hence Borel.

Now

and a countable union of Borel sets is Borel. Therefore .

The theorem used

The Borel -algebra is closed under countable unions. Since singletons are closed sets, and closed sets are Borel, any countable set is Borel.

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Question 10. Prove that any nonempty open subset of is a countable union of open intervals, in fact disjoint. The same is true for nonempty open sets of with intervals replaced by sets of the form . However, as in , they may not be disjoint anymore.

Answer for

Let be nonempty and open.

For each , consider the set of points in that can be connected to by an interval staying inside :

Because is open, each is an open interval. Two such intervals are either equal or disjoint, so they partition into disjoint open intervals.

To see that there are only countably many of them, note that every nonempty open interval contains a rational number. Distinct disjoint intervals contain distinct rationals, so the collection is at most countable.

Therefore is a countable disjoint union of open intervals.

A more concrete proof

Equivalently, one may define the maximal open interval in containing each . These maximal intervals are disjoint, open, and their union is exactly .

Answer for

Let be nonempty and open.

For every , because is open there exists an open box

such that

Using rational endpoints and the density of in , we can choose boxes with rational endpoints inside . There are only countably many such boxes, because is countable.

Hence is a countable union of open boxes.

Unlike the one-dimensional case, these boxes need not be disjoint.

The theorem used

The density of the rationals says that every open interval contains a rational number. This is what makes the one-dimensional disjoint decomposition countable.

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Question 11. For the following collections of subsets , , of prove that .

Answer

We show that each collection generates the Borel -algebra.

(a)

Every set in is open, so it is Borel. Hence

Now let be an open interval. Then

Each set on the right is in because and are in , and -algebras are closed under complements and intersections.

Therefore every open interval is in , so every open set is in , and hence

Thus .

(b)

Because is dense in , for any real we have

So every set in belongs to , which gives

But from part (a), . Also , so .

Therefore .

(c)

Each rational open interval is Borel, so

Conversely, every open interval can be written as a countable union of rational open intervals:

Thus every open interval is in , hence every open set is in . Therefore

So .

The key theorem used

The Borel -algebra on is generated by any one of the standard bases for the topology: open rays, open intervals, or open intervals with rational endpoints.

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Question 12. For the following collections of subsets , , of , , prove that .

Answer

We use the fact that the Borel -algebra on is generated by open boxes with rational endpoints.

(a)

Every open box belongs to , so the -algebra generated by contains all open boxes. Since open boxes generate the topology on , the generated -algebra contains all open sets, hence contains .

Conversely, every element of is a Borel set, because it is a finite product of Borel intervals. Therefore .

So .

(b)

For each coordinate and real number , we can write

for rationals .

Thus the real-endpoint rays in are generated from rational-endpoint rays in , so

On the other hand, every set in is Borel, so .

Since is rich enough to generate all open boxes, it follows that .

(c)

The collection is countable, and its elements are Borel. Hence .

Conversely, because rational points are dense in , every open box contains a rational box, and every open set is a countable union of rational boxes. Therefore .

Thus .

(d)

The family consists of boxes with rational endpoints, allowing infinite endpoints as well. These are all Borel, so the generated -algebra is contained in .

But rational open boxes already generate all open sets, so .

Therefore .

The theorem used

In , the Borel -algebra is generated by open boxes, and the rational boxes form a countable base for the topology.

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Question 13. If then prove that and is properly contained in .

Answer

Here is the collection of all singleton subsets of .

We show that the -algebra generated by all singletons is exactly the countable/co-countable -algebra.

Step 1: Every countable set is in

Since each singleton lies in , and -algebras are closed under countable unions, every countable union of singletons belongs to .

Therefore every countable subset of belongs to .

Step 2: Every co-countable set is in

If is countable, then by Step 1. Since is closed under complements, is also in .

So every co-countable set belongs to .

Thus

Step 3: Nothing else is needed

Let be any -algebra containing all singletons. Then it contains all countable unions of singletons, hence all countable subsets of . By closure under complements, it also contains all co-countable sets. So for every such .

Therefore is the smallest -algebra containing , which means

Proper containment in

To show , it is enough to exhibit a Borel set that is neither countable nor co-countable. Any nonempty open interval works, for instance .

The interval is Borel, but it is uncountable and its complement is also uncountable, so it does not lie in .

Hence is properly contained in .

The key theorem used

Any -algebra containing all singletons must contain all countable unions of singletons, and therefore all countable sets. Taking complements gives all co-countable sets.

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Question 14. Let be the extended real line. The Borel -algebra on , denoted by , is defined as the smallest -algebra containing . Prove that

Answer

We prove that the right-hand side is a -algebra containing and the two endpoints, and then use minimality.

Step 1: The displayed family contains and the endpoints

If , then taking gives .

Also,

So the family contains .

Step 2: Closed under complements

Let with and .

There are four cases, depending on whether and belong to .

• If , then .
• If , then .
• If , then .
• If , then .

In every case, is Borel and the endpoint part is again a subset of .

So the family is closed under complements.

Step 3: Closed under countable unions

If with and , then

The union is Borel, and is still a subset of .

Hence the family is a -algebra.

Step 4: Conclude minimality

This -algebra contains together with the two points , so it contains the -algebra generated by them. Conversely, any -algebra containing and the endpoints must contain every set of the displayed form.

Therefore

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Question 15. Let and be nonempty sets and be a given function. If , define

(even if the inverse function does not exist). If is a -algebra on then prove that

is a -algebra on .

Answer

We use the standard properties of inverse images:

and

Now we check the axioms.

(i) and belong to

Since and ,

So .

(ii) Closed under complements

If , then for some . Hence

Because , we get .

(iii) Closed under countable unions

If with , then

Since is a -algebra, , so the union belongs to .

Therefore is a -algebra on .

The theorem used

Inverse image preserves arbitrary unions, complements, and therefore all -algebra operations.

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Question 16. The aim of this exercise is to prove that any -algebra is either finite or uncountable.

(a) Suppose is a nonempty set and is a -algebra on . If is an infinite set prove that so is .

Answer

We prove the contrapositive.

If is finite, then its power set is finite, because a finite set has only finitely many subsets. Since every -algebra on is a subset of , it must also be finite.

Therefore, if is infinite, then cannot be finite. Hence is infinite.

(b) Suppose is a countably infinite set and so is . Define by

Prove that is the smallest set in containing .

Answer

For a fixed , consider the family of all sets in that contain . Their intersection is exactly .

We must show two things:

1. .
2. If and , then .

Why

Since is countable, the family of sets in containing is countable. A countable intersection can be rewritten using complements as

Because is closed under complements and countable unions, the right-hand side lies in .

So .

Why it is the smallest such set

If and , then is one of the sets being intersected. Therefore the intersection is contained in :

And since belongs to every set in the intersection, we also have .

Thus is the smallest member of that contains .

(c) If , , and , then prove that . Conclude that (which is a subset of ) is a partition of .

Answer

Suppose , and choose .

Since is the smallest measurable set containing , and , we know that . Similarly, because , we get .

But also , so by minimality of among sets containing and using the fact that , we can reverse the inclusion argument to obtain equality. More simply: two minimal sets in a -algebra containing points from a common intersection must coincide.

Hence

Therefore distinct sets in are disjoint. Since every point belongs to its own set , the union of all sets in is .

So is a partition of .

(d) If then prove that

Answer

Take any . Since is a measurable set containing , and is the smallest measurable set containing , we must have

So

Conversely, if , then , and since is one of the points indexing the union, we get .

Hence

(e) If is an infinite set then prove that is an infinite subset of .

Answer

This part is understood in the context of the exercise where is assumed infinite.

If were finite, then by part (d), every set in would be a union of some subcollection of the finitely many atoms in . That would give only finitely many possible unions, hence only finitely many sets in .

But is assumed infinite. Therefore cannot be finite.

So is infinite.

(f) If is an infinite set and is an infinite set then must be uncountable.

Answer

Assume, for contradiction, that is countably infinite.

Then by the construction in parts (b) through (e), the set of atoms is infinite. But is countably infinite, so the partition can have only countably many atoms, and each measurable set is a union of atoms.

Now part (d) says every measurable set is a union of some subcollection of the atoms. If there are infinitely many atoms, then there are at least as many measurable sets as subsets of a countably infinite set of atoms. The power set of a countably infinite set is uncountable.

So cannot be countably infinite. Since we already know from part (a) that an infinite forces infinite, the only possibility is that is uncountable.

Final conclusion

Every -algebra is either finite or uncountable.

Main theorem behind the argument

The atoms of a -algebra form a partition of the space, and every measurable set is a union of atoms. A countably infinite collection of atoms would produce uncountably many unions, so an infinite -algebra cannot remain countable.