Checking whether a finite family forms a sigma algebra

Key concept: This exercise is about verifying the three defining properties of a sigma algebra: containing the whole set, being closed under complements, and being closed under countable unions.

Step by step

What a sigma algebra must satisfy

A collection $\mathcal A$ of subsets of a set $X$ is a $\sigma$-algebra if it contains $X$, is closed under complements relative to $X$, and is closed under countable unions. For finite examples, checking complements and unions is usually enough.

A quick practical test is: if one set is in the family, then its complement must also be there. Then any union of sets already in the family must again belong to the family. If even one complement is missing, the collection is not a $\sigma$-algebra.

How to check each finite collection

For each option, compare every subset with its complement in $X=\{1,2,3,4,5,6\}$.

• In (a), $\{1\}$ is present, but its complement $\{2,3,4,5,6\}$ is not listed, so it fails.
• In (b), $\{1,3\}$ and $\{2,4,5,6\}$ are complements of each other, and the collection contains $\varnothing$ and $X$, so this one works.
• In (c), $\{1\}$ is present, but its complement $\{2,3,4,5,6\}$ is also present, and the other listed sets pair correctly; however, you still must check closure under unions of all listed subsets. The family is designed to behave like a generated partition-based $\sigma$-algebra.
• In (d), several complements are missing or mismatched, so it fails.

Why complement closure is the fastest test

When the universe is finite, a $\sigma$-algebra is often built from a partition of $X$. Every set in the algebra is a union of blocks of that partition. That means the family automatically contains complements and unions. If the listed subsets do not match that structure, the collection is usually not a $\sigma$-algebra.

Key property to remember

The presence of $\varnothing$ and $X$ is necessary but not sufficient. A family can include both and still fail because a complement is missing or because a union of two listed sets is absent. That is why checking the full closure conditions matters.

Pitfall alert

A common mistake is to see $\varnothing$ and $X$ in the family and immediately assume it is a sigma algebra. Those two sets are required, but they are not enough. Another error is checking only a few complements instead of every listed subset. Since a sigma algebra must be closed under complements, one missing complement is enough to reject the collection. For finite universes, it also helps to think in terms of partitions: if the family is not generated by unions of partition blocks, it is unlikely to qualify. Do not forget that closure under unions must also hold, not just complements.

Try different conditions

If the universe changed to $X=\{1,2,3,4\}$ and the family were $\{\varnothing,\{1,2\},\{3,4\},X\}$, then it would be a sigma algebra because each set has its complement in the family and unions stay inside the family. But if we added $\{1\}$ to that same family without adding $\{2\}$, it would immediately stop being a sigma algebra. A small change in one subset can break complement closure, which is why these questions are often about structure rather than listing every set individually.

Further reading

complement closure, countable union, partition-generated sigma algebra