Indicator function confusion

Hi,

I'm currently reading a book on probability theory (Resnick, "A Probability Path") and in a section on basic set and measure theory, the indicator function is defined as follows:

and some properties are discussed, including:

for which it is stated: "and if the sequence is mutually disjoint, then equality holds."

However, I don't understand this: the LHS can only ever be 0 or 1, whereas the RHS can be any integer between 0 and the total number of subsets. How could equality ever hold except under specific cases?

**Update:** Actually, I just realised I'm being stupid. If the sets are disjoint, then the RHS would also only return 0 or 1.

Any pointers greatly appreciated!

Chris

Re: Indicator function confusion

Quote:

Originally Posted by

**entropyslave**

What is the definition of the infimum of a family of sets? Does it mean intersection?

Quote:

Originally Posted by

**entropyslave** There is a proof given for this property that is based on proving the LHS is 1 iff for some element

,

for all

which is also true for the RHS. However, I don't understand how this equality holds since if

for some but not all

, then the RHS will be the null set (whereas the LHS is only 0 or 1).

The RHS is a number, not a set. I assume that there is a universal set so that for all . Then if is a family of functions from to (or from to {0, 1} in this case), is probably defined pointwise: for every . So, , and it equals 1 iff for all .

Quote:

Originally Posted by

**entropyslave** Another relation introduced is:

for which it is stated: "and if the sequence

is mutually disjoint, then equality holds."

However, I don't understand this: the LHS can only ever be 0 or 1, whereas the RHS can be any integer between 0 and the total number of subsets. How could equality ever hold except under specific cases

It does hold under specific cases: when the sequence is mutually disjoint. The left-hand side equals 1 when belongs to any of , whereas in the right-hand side, you add 1 for each to which belongs.

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