Question: [Extensions of measurable functions](https://math.stackexchange.com/q/76568)
Let $(X,F)$ be a measurable space. $U \subset X$, which is not necessarily measurable. $f$ is a real-valued measurable function defined on $(U,F\cap U)$, then we can extend $f$ to a measurable function on $X$.
The answer in the link above makes me confused, I don’t know because $U$ must be measurable.

Answer: 
To prove (1), please look up [Bolzano-Weierstrass theorem](https://en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem)

For (2), note that any subsequence of your original bounded sequence is going to be bounded as well. Therefore, we know such subsequence will, itself, contain a converging subsequence (by (1)). Since there are an infinite number of distinct subsequences of the original sequence (if the sequence is not constant after a certain index), there are an infinite number of converging subsequences.
Mere bhai , you can only do this , select a element and repeat it infinite times only when range set is finite.

The idea of this proof is that , if a sequence is bounded then it's means it's range set is bounded . And you may also know about that every bounded subset of $\mathbb{R}^k$ lies in compact subset of $\mathbb {R}^k$ and its a fact that sequence in compact matric space must have a sub sequence in it.