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$f=I$

The first function is not continuous.(no Lebesgue integrable)

The condition 2 is not true.
The sequence must be decreasing

Why there exist $a,b\in S$ with $(a,b)=1$?

What if $ \int _{A}g=0$ and $ \int _{A}fg\neq 0 $ ?

It is not bounded near zero.

We have
1)Every finite extension is algebraic
2)The algebraic closure of real numbers are
the complex numbers
3)Every finite extension of real numbers are
the complex numbers.

1) subspace of a vector space for me( Rudin Functional Analysis and other) mean linear subspace.
2)This is a simple version of Tietze Extension Theorem.










0

If we restrict the function to the half-plane where ${\rm Re}(z) \geq 0$ , then $g$ is bounded which by a known fact of complex analysis means it is constant. The above is false. $g(z)=\exp(-z)$.

Is easy to prove:
The range of entire no constant function is dense.

1) If X is compact then is not subspace
2)The functions in Hahn-Banach theorem are linear