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Is it true that for \(a,b\geq 0\) and \(f,g\geq 0\) we have
Solution: No, only for \(p=1\) or for \(g=0\). But it remains true as an inequality for all \(1\leq p\leq ∞\).
Is it true that for \(F,E\) disjoint we have
Solution: No, only for \(p=1\).
Is it true for \(0\leq f\leq g\) that \(\|f\|_{L^p(ℝ^d)}\leq \|g\|_{L^p(ℝ^d)}\)?
Solution: Yes.
Does \(\|f\|_{L^p(ℝ^d)}=0\) imply \(f=0\) everywhere?
Solution: No, only almost everywhere.
What is another way to write \(\|f\|_{L^∞(ℝ^d)}\) if \(f:ℝ^d→ℝ\) is continuous?
Solution: If \(f\) is continuous then \(\|f\|_{L^∞(ℝ^d)}=\sup _x|f(x)\).
If \(E⊂ℝ^2\) is \(\lmgd 2\)-measurable, is it true that for each \(x∈ℝ\) the set \(E_x=\{y∈ℝ:(x,y)∈E\}\) is \(\lmgd 1\)-measurable?
Solution: No (assuming AC). For example take a nonmeasurable set \(S⊂[0,1]\) and set \(E=\{(0,y):y∈ S\}\). Then \(\lmgd 2(E)=0\) and thus \(E\) is \(\lmgd 2\)-measurable, but \(E_0=S\) is not \(\lmgd 1\)-measurable.
If \(f:ℝ^2→[-∞,∞]\) is \(\lmgd 2\)-integrable, is it true that for every \(x∈ℝ\) the function \(y↦f(x,y)\) is \(\lmgd 1\)-integrable?
Solution: No, only for \(\lmgd 1\)-almost every \(x∈ℝ\). For example \(f=∞\cdot (\ind {\{0\}×[1,2]}-\ind {\{0\}×[-1,0]}\) has \(∫f\intd \lmgd 2=0\) but \(y↦f(0,y)\) is not \(\lmgd 1\)-integrable.