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Let \(A_1,A_2,…⊂Ω\) be measurable and \(a_1,a_2,…\geq 0\). Is the function \(f=\sum _{n=1}^∞a_n\ind {A_n}\) measurable?
Solution: Yes, because for each \(k∈ℕ\) the function \(f_k=\sum _{n=1}^ka_k\ind {A_k}\) is measurable and \(f=\lim _{k→∞}f_k\).
Let \(a_1,…,a_n,b_1,…,b_n∈ℝ\) and \(A_1,…,A_n,B_1,…,B_n⊂Ω\) such that
Solution: No, it may happen that \(A_1=A_2=B_1=B_2\) and \(a_1+a_2=b_1+b_2\) but \(\{a_1,a_2\}\neq \{b_1,b_2\}\), for example for \(a_1=0,a_2=2,\ b_1=b_2=1\).
It similarly happen that \(a_1=a_2=b_1=b_2\) and \(A_1∪A_2=B_1∪B_2\) but \(\{A_1,A_2\}\neq \{B_1,B_2\}\).
Let \(f_1,f_2:Ω→[0,∞]\) and \(f=f_1-f_2\). Does that imply \(f_1=f^+\) and \(f_2=f^-\)?
What, if for all \(x∈Ω\) we have \(f_1(x)=0\) or \(f_2(x)=0\)?
Solution: No, for example if \(f=1\) is constant, then \(f^+=1\) and \(f^-=0\) but \(f_1=2\) and \(f_2=1\) satisfy \(f=f_1-f_2\).
The statement is true however under the assumption that for all \(x∈Ω\) we have \(f_1(x)=0\) or \(f_2(x)=0\).
Let \(μ\) be a measure on \(ℝ^d\). Is \(ℝ^d\) \(σ\)-finite with respect to \(μ\)?
What, if \(μ\) is a Radon measure?
Solution: No, for example for the counting measure.
If \(μ\) is a Radon measure the answer is yes, because \(ℝ^d=B(0,1)∪B(0,2)∪…\) and \(μ(B(0,n))\leqμ (\tc {B(0,1)})<∞\).
Given \(f,f_1,f_2,…:Ω→ℝ\), what does it mean that \(f_n→f\) uniformly?
Solution: It means that for every \(ε>0\) exists an \(n∈ℕ\) such that for all \(k\geq n\) and \(x∈Ω\) we have \(|f_k(x)-f(x)|<ε\).
Find an example of functions \(f,f_1,f_2,…:(0,1)→ℝ\) such that \(f_n→f\) pointwise, but not uniformly.
Solution: Set \(f_n(x)=x^{1/n}\). Then \(f_n→0\) pointwise but not uniformly: Let \(n∈ℕ\). Then for \(x_n=1/2^{1/n}∈(0,1)\) we have \(|f_n(x_n)-0|=1/2\).
Find \(f:ℝ→ℝ\) and a closed set \(C⊂ℝ\) such that \(f:C→ℝ\) is continuous but \(f:ℝ→ℝ\) is not continuous in every \(x∈C\).
Can you also find an open set with that property?
Solution: Set \(C=[0,1]\) and \(f(x)=1\) if \(x∈C\) and \(f(x)=0\) if \(x\notin C\).
Such an example cannot be found with an open set \(U\) instead. The reason is, that if \(x∈U\) and \(x_n→x\), then there exists a \(k∈ℕ\) such that for all \(n\geq k\) we have \(x_n∈ U\). As a consequence, \(f:U→ℝ\) is continuous in \(x\) if and only if \(f:ℝ^d→ℝ\) is continuous in \(x\).
What is the Lebesgue integral \(∫f\intd \lmg \) for
Solution: