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### $\| f \|_{L^1} > 0$ is stronger than simply $f \neq 0$. But how much?

Yes; if $\|f\|_p>0$ then $f>0$ on some positive measure set, since otherwise $f=0$ a.e. and hence $\|f\|_p = 0$. Conversely, if $\|f\|_p = 0$ then $f=0$ almost everywhere.
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### Product of a $L^{p}$ function and measure

Your intuition is correct. You cannot multiply a 'function' in $L^p(\nu)$ with a measure $\mu$ unless $\nu$ is absolutely continuous w.r.t. $\mu$. This is independent of summability issues and just a ...
1 vote
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By definition $$p_{-}=\sup\{a\,:\,\mu\{x\,:\, p(x)<a\}=0$$ If $p_{-}=\infty$ then for every $a$ there holds $\mu\{x\,:\, p(x)<a\}=0.$ We have $$\{x\,:\, p(x)<\infty\}=\bigcup_{n=1}^\infty \{x\... -2 votes ### Is a set of representatives of \Bbb R/\Bbb Z[\frac12] a Vitali set? Every set of representatives of \Bbb R/\Bbb Z[\frac12] will contain a pair of elements differing by a rational not in \Bbb Z[\frac12] e.g. \frac13 and therefore it cannot be a Vitali set. ... 1 vote ### Find Lebesgue measurable sets such that the measure of intersection is product of measures. Denote the partition of (0, 1) into 2^n equal intervals by P_n. Let A_n be the union of the odd numbered intervals in P_n. The Lebesgue measure of A_n is 1/2. Now suppose n > m. ... 4 votes Accepted ### Find Lebesgue measurable sets such that the measure of intersection is product of measures. First of all, A_n = \emptyset or (0,1) for all n works but I guess you want a non trivial example. The equality m(A_n \cap A_k) = m(A_n)m(A_k) means, from the probabilitic point of view, that ... 0 votes ### What is the Lebesgue measure of Luzin's non-Borel set of reals? There exists a result for continued fractions similar of "normal" numbers: for almost every x\in [0,1], the number k occurs in the continued fraction of k (in the limit) with ... 2 votes Accepted ### What is the Lebesgue measure of Luzin's non-Borel set of reals? I now believe justt's conjecture that the Lebesgue measure of A is one, and I believe I have a proof. It uses some similar ideas to justt's proof outline (in particular, Khinchin's constant), but in ... 1 vote Accepted ### Understanding proof of "if f is non-negative, then it is the supremum of simple functions." For your first question, yes that is true, but I am not sure why that would be helpful in proving \int_B h \, d\mu \le \sup\{\cdots\}. For your second question: This follows from the definition of ... 2 votes Accepted ### Optimal Transport Theory: Find a sequence of functions that converge weakly to 1/2 For (a), under the \pi-\lambda Theorem, it suffices to prove \mathcal{L}_{[0,1]}(T_n^{-1}([a,b]))=\mathcal{L}_{[0,1]}([a,b]) for all intervals [a,b]\subset[0,1], n\in \mathbb{N} (here \... 3 votes Accepted ### Brezis' Proposition 8.4: an inequality that holds for a.e. x,y \in \mathbb R Your proof is OK. Maybe an easier proof is to show that$$A:=\{x\in\mathbb R: x\in N \text{ or } x+h\in N\} has measure zero, hence for $\lambda\text{-a.e. }x\in\mathbb R$ we have $x\in N^c$ and $x+... 1 vote Accepted ### What does it mean that a random variable has "Lebesgue density"? Lebesgue density will refer to a "density" with respect to the Lebesgue measure. If you don't know much measure theory, then just read this as: "$Y$has a probability density" (a ... 3 votes Accepted ### Pointwise a.e. approximation by a sequence of smooth functions with supremum bound The trick here is to use bump functions. Specifically, we are using the fact that if$K\subseteq U$with$K$compact and$U$open, and$c_1$and$c_2$are constants, then there is a smooth (i.e., in$...
No, I do not think it is. And if by your observation that $u$ being square integrable implies it being finite almost everywhere you mean that it must be bounded (i.e., $|u(x,t)|\leq k$ for some $k$) ...