Thomas E.
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 Jun5 comment Limit of $\frac{x^{x^x}}{x}$ as $x\to 0^+$ Maybe the last equality should be $...=-2\cdot 0\cdot 1=0$ instead? Jun4 comment Continuous function on metric space @JasonDeVito: You probably mean that the sum equals $0$? Jun4 comment Continuous function on metric space @Hiperion: Disjointess is not enough (but it is required), you have to use the property that $A$ and $B$ are closed. Hint: $\bar{A}=\{x\in X:d(x,A)=0\}$, where $\bar{A}$ is the closure of $A$ in $X$. Jun4 revised Continuous function on metric space edited tags Jun4 comment Checking for completeness of $\mathbb{R}$ with metric defined by $d_1(x,y) =\mid e^x - e^y \mid$ @srijan: yes, that's correct. Jun4 comment Point set topology question: compact Hausdorff topologies You probably mean "tau", which is \tau and looks the following: $\tau$. Jun4 comment Checking for completeness of $\mathbb{R}$ with metric defined by $d_1(x,y) =\mid e^x - e^y \mid$ For $d_{1}$, note that you must start with an arbitrary Cauchy sequence in $(\mathbb{R},d_{1})$ and not in the standard metric. The current argumentation does not show that $d_{1}$ is complete. Jun4 comment $\varphi\colon M\to N$ continuous and open. Then $f$ continuous iff $f\circ\varphi$ continuous. How does this proof show that $f^{-1}A$ is open in $N$? Jun4 revised Sequential characterization of closedness of the set edited tags Jun4 comment Almost sure convergence for sequence of function If the sequence $(f_{t})$ is indexed by $t\in (0,1)$, how can you observe limiting when $t\to\infty$? Jun3 comment Sequential characterization of closedness of the set @xan: Sure; you're welcome. I'm glad I was able to help you figure this out. Jun3 answered Sequential characterization of closedness of the set Jun3 comment Showing a subset of $C([0,1])$ is compact. What topology does $C([0,1])$ have? The uniform one? If yes, then work with sequences instead of covers. Try showing that any sequence has a uniformly convergent subsequence. Jun1 comment Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral Sure; I corrected it and re-opened the answer. Jun1 revised Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral added 161 characters in body Jun1 comment Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral ... by taking sup over all such $s$ we obtain $\int_{\mathbb{R}} f\leq \int_{A}f+\int_{A^{c}}f$. Can you show the other direction similarly? By choosing arbitrary simple functions $0\leq s_{1}\leq f|_{A}$ and $0\leq s_{2}\leq f|_{A^{c}}$, and showing $\int_{A}s_{1}+\int_{A}s_{2}\leq \int_{\mathbb{R}}f$, and taking supremum over all such $s_{1}$ and $s_{2}$ obtaining the other inequality. Jun1 comment Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral With this lemma it works. If you want to avoid working suprumems over sets in equalities, you may do something like the following too (you already got the idea). $"\Rightarrow"$: Let $0\leq s\leq f$ be arbitrary simple function, and define $s_{1}=s 1_{A}$ and $s_{2}=s 1_{A^{c}}$, whence $s=s_{1}+s_{2}$ and $s_{1}(x)\leq f(x)$ for $x\in A$ and $s_{2}(x)\leq f(x)$ for $x\in A^{c}$. Now using what you had already proven $\int_{R} s=\int_{R} s 1_{A}+s 1_{A^{c}}=\int_{R}s 1_{A}+\int_{R}s 1_{A^{c}}=\int_{A} s_{1}+\int_{A^{c}} s_{2}\leq \int_{A}f+\int_{A^{c}}f$. (Continues below) Jun1 comment Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral There was a tiny flaw, which I should've fixed, and I saw you posted your own:) I will read it soon. Jun1 answered Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral Jun1 comment Proof of $\int\limits_{A}f=\int\limits_{\mathbb{R}}f{1}_A$ for the Lebesgue integral Have you proved that for any non-negative measurable function $f$ there exists a nondecreasing sequence of simple functions $(\psi_{i})$ so that $\psi_{i}\to f$ point-wise? With this and monotone convergence theorem you may conclude the result. Or you may want to notice that $\int_{A} f=\int_{A}f+\int_{A^{c}}0=\int_{A}f1_{A}+\int_{A^{c}}f1_{A}=\int_{\mathbb{R}}f 1_{A}$.