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 3h revised Set all measurable real functions on $[0,1]$ with metric $\int_{0}^1 \min \{1,|f(t)−g(t)|\}dt$ is Fréchet without nonzero continuous linear functional deleted 36 characters in body 5h comment Why does the Dominated Convergence Theorem fail for evaluating $\lim_{n \to \infty} \ n^2 \int_{0}^{1} xe^{-n^2x^2}dx$? @mickep Indeed the OP probably studied $\sup\limits_{n\geqslant1}nxe^{-nx^2}$ instead of $\sup\limits_{n\geqslant1}n^2xe^{-nx^2}$. Note that even $\sup\limits_{n\geqslant1}nxe^{-nx^2}$ is not exactly $1/(xe)$ (except when $1/x^2$ happens to be an integer) since $n$ is restricted to be an integer, in fact $\sup\limits_{n\geqslant1}nxe^{-nx^2}\leqslant1/(xe)$. Note finally that this last inequality would not be sufficient to conclude (if the question was about the convergence of the integral of the functions $nxe^{-nx^2}$) since an integrable function can be bounded above by a nonintegrable one. 5h comment How do you do obtain the sum $244 + 132$ in base $5$? @CiaPan "Fallacious" is too strong. I was not aware of the approach using such intermediary steps when I discovered this answer but, after having thought about it, I fail to see what could go wrong using them. Maybe you could be more specific about it? 6h comment How do you do obtain the sum $244 + 132$ in base $5$? At least the result is correct (the only one on the page, so far). 6h comment Why does the Dominated Convergence Theorem fail for evaluating $\lim_{n \to \infty} \ n^2 \int_{0}^{1} xe^{-n^2x^2}dx$? If $x=1/k$ for some positive integer $k$ and $n=2k^2$, then $n^2xe^{-nx^2}=4k^3e^{-2}=4e^{-2}/x^3$ hence the value of the supremum that you give is not exact (and not of the correct order, which is $\Theta(1/x^3)$ when $x\to0$). 7h revised Expectation of a quotient of random variables edited body; edited tags 7h comment Expectation of a quotient of random variables $$E(X)=E(E(X\mid Y))=E\left(\tfrac12Y\right)=\int_0^1\tfrac12y\cdot2y\,dy=\ldots$$ 7h comment Expectation of a quotient of random variables $$E(XY^{-1})=E(E(XY^{-1}\mid Y))=E(E(X\mid Y)Y^{-1})=E\left(\tfrac12Y\,Y^{-1}\right)=\ldots$$ 8h comment Equality of integrals and Almost sure equality of random variables There is slightly more to this when $\mathcal D$ is a strict sub-sigma-algebra of $\mathcal F$, see my comment on main. 8h comment Equality of integrals and Almost sure equality of random variables Sufficient condition to conclude that $P(X=Y)=1$: that $X-Y$ is $\mathcal D$-measurable. Reverse implication: yes, if $P(X=Y)=1$ then $E(X;D)=E(Y;D)$ for every $D$ in every sub-sigma-algebra $\mathcal D$ of the sigma-algebra $\mathcal F$. 8h comment Limit definition for big O Of course it is. What is your point exactly? 8h comment Find the domain for which the function $f(z)=\int\limits_{-1}^{1}\frac{e^{tz}}{1+t^2}dt$ is defined and holomorphic? Hint: $$f(z)=\sum_{n=0}^\infty a_n\frac{z^n}{n!}\qquad a_n=\int_{-1}^1\ldots dt,$$ hence, for every $n$, $$|a_n|\leqslant\ldots$$ which proves that the radius of convergence is $_____$. 8h comment Cauchy Criterion - Prove $\sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^k}{\sqrt k}$ converges uniformly on $[0,1]$ @robjohn Three rather contorted comments to basically confirm the point I made seem more than enough. 8h comment How to define a bijection between $(0,1)$ and $(0,1]$? @ToddWilcox Did I mention votes? 9h comment Limit definition for big O Reviving this question from two years ago to post this very partial answer was not needed. Note that, in the restricted case when $g(n)\ne0$ for every $n$ large enough, $$f(n)\in O(g(n))\iff\limsup_{n\to\infty}\left|\frac{f(n)}{g(n)}\right|<+\infty.$$ 9h comment Conditional version of bernstein's inequality Unless you show the "unconditional" proof, it is difficult to say if this is indeed a problem or not (I suspect it is not). 10h comment algebraic derivation involving integration , expectation, and max function Hint: This is a pointwise identity, that is, getting rid of the irrelevant factor $L_0$, for every real numbers $(k,r_0,r)$, one has $$(k-r_0)(r-k)^++2\int_k^\infty (r-x)^+dx=(r-r_0)(r-k)^+.$$ The proof is direct, treating $rk$ separately. (Unrelated: $k$ in the LHS of the identity in your question should read $K$.) 10h comment $A$ closed, $\mathbb{R^n} \setminus A$ open? Missing "boundary points". (You know, we should not have to insist like that.) 10h comment Convergence in $L^p$ and almost sure convergence Yes 3 holds since $|X|^j\leqslant1+|X|^k$ pointwisely. Re 4, indeed $L^p$ convergence implies convergence in probability and almost sure convergence implies convergence in probability. Some partial reverse implications are that convergence in probability implies almost sure convergence of a subsequence and that convergence in probability plus some uniform integrability imply $L^p$ convergence. (Fully agree with the "this is standard stuff" remarks.) 10h comment Conditional version of bernstein's inequality Such inequalities often follow from integrating pointwise inequalities between random variables. Did you check whether integrating the conditional versions of these pointwise inequalities was not enough? (Unrelated: In the versions of Bernstein's inequality that I know, one assumes that \$|X|