# Can a nonempty set has probability zero?

Reworded version:

I think I need to re-word as follow:

Can a nonempty set X with P(X)=0?

Suppose F to be a non-empty set with $$P(F)\neq 0$$. Call its closure be F'.

Now let set theoretic different F'\F be A. Clearly, A could be non-empty.
Is this the case where P(A)=0?

Is not, how do you relate P(F) and P(F')?

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Original version:
Can a nonempty set X has P(X)=0?

My thought is suppose a nonempty set F then its closure F'\F=A where P(A)=0.

Is this true??

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lavinia
Gold Member
Can a nonempty set X has P(X)=0?

My thought is suppose a nonempty set F then its closure F'\F=A where P(A)=0.

Is this true??
No. The closure of a set of probability 0 could have probability 1

No. The closure of a set of probability 0 could have probability 1
I think I need to re-word as follow:

Suppose F to be a non-empty set with $$P(F)\neq 0$$. Call its closure be F'.

Now let set theoretic different F'\F be A. Clearly, A could be non-empty.
Is this the case where P(A)=0?

Is not, how do you relate P(F) and P(F')?

Last edited:
Homework Helper
Think of an absolutely continuous distribution (Gaussian as one example). The interval (0,1) could have non-zero probability. Its closure is [0,1], also with non-zero probability, but the difference you reference is $$\{0,1\}$$ which has zero probability.

No. The closure of a set of probability 0 could have probability 1
Of course: throw a dart at the real line. Then P(hitting a rational)=0, but

P(hitting some element in Cl(Q))=1

Then this is also an example of a set being non-empty, yet having probability zero.

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Thanks for those commented above.
Now I understand that the probability of a set and probability its closure equals depends on the probability measure (or probability mass function) and of course the set itself.