For epsilon and outer Measure this inequality holds

Please Donate Money ('' Shagun ka ek rupay'') for this Channel

Suppose we have a non empty set E which is subset of R this implies that E is a set which contains real numbers. By the definition of outer measure of E we have to take a Infimum of real numbers which are obtain by the sum of lengths of open bounded interval of each family which covers E called this set A so there outer measure of E is Infimum of A.
By this result
We can successfully write the required inequality
Which is used in the proof of Outer measure is countably subaddtive.

[harshvardhansinghbhoj]

One thing i dont understand in this proof, in the claiming part you use <= but when you proved it you use = so how they both are same ??

[wannnabee]

Which book do you prefer for measure theory sir?

[NDjayswal]

When you say sequence of sets, its not it.
{Aᵢ} means {A₁, A₂, A₃, ...} it is a countable collection of sets.
also this is countable so it can be written as sequence 〈Aᵢ〉.
but we not using sequence we using countable collection of sets.
{Aᵢ} is not even set, so we must say only collection of Aᵢ's, i∈ℕ.

[paramaguru]

A=(0, 1)..A Belongs to B=(0, 10).. outer measure of A is 1..
10 doesn't less