# chango

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 Apr12 comment Given $L(y) = \int_ {1}^{y}\frac{1}{t}dy$ , $y\in(0, \infty)$$n$ starts at $2$ in the last sum... Apr12 answered Does the closed graph theorem presuppose that the domain is closed? Apr12 comment Given $L(y) = \int_ {1}^{y}\frac{1}{t}dy$ , $y\in(0, \infty)$yes it does. it ensures differentiability which is stronger. Apr12 comment Egorov's Theorem - Counterexample in Infinite CaseHere your hypothesis of finite measure fails since $m(\mathbb{R}) = \infty$ and the conclusion fails as well because for any set of finite measure $F$, your function won't converge uniformly to the zero function. Apr12 comment Egorov's Theorem - Counterexample in Infinite CaseThe definition states that if you have a sequence of functions in a set of finite measure $E$ which converge pointwise to $f$ almost everywhere. Then for every $\epsilon >0$ you can find a measurable set $F \subset E$ such that $m(F) < \epsilon$ and $f_n$ converges uniformly to $f$ on $E \setminus F$. Apr11 answered Egorov's Theorem - Counterexample in Infinite Case Apr11 comment Given $L(y) = \int_ {1}^{y}\frac{1}{t}dy$ , $y\in(0, \infty)$you basically have it. the FTC tells you that your function is differentiable (and therefore continuous) at all $x \in (0,\infty)$. The increasing bit you get (as mentioned by Mark) by noting that the integral of a positive function over an interval gives you a positive number. Let $z > y$ be two numbers in $(0,\infty)$. Then $L(z) - L(y) = \int_y^z \frac{1}{t} dt > 0$ Apr11 comment Given $L(y) = \int_ {1}^{y}\frac{1}{t}dy$ , $y\in(0, \infty)$And forgot to mention, you also multiply by the derivative with respect to $y$ of the upper bound function, in your case this is $f(y)= zy$ so you get the $z$ in the numerator. It's basically chain rule. Apr11 comment Given $L(y) = \int_ {1}^{y}\frac{1}{t}dy$ , $y\in(0, \infty)$The first and the second term appear because you need to use the fundamental theorem of calculus for differentiating an integral. Basically you evaluate the integrand at the endpoints... Apr11 comment Equivalence of measures and $L^1$ functionsgot that now. thanks! Apr11 comment Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2$.No worries. Good luck! Apr10 comment Let $f:\Bbb R^2→\Bbb R:(0,0)\mapsto 0 \quad (x,y)\mapsto\frac{x^2y^2}{x^4+y^4}$Yeah, that is the right idea: $\lim_{x \to 0} f(x,x) = 1/2 \neq 0 = \lim_{x \to 0} f(x,0)$ Apr10 comment Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2$.Also, I think you should be less rude to people who are trying to help you. And be humble when you are asking for help, advice, etcetera. Apr10 comment Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2$.Oh well, whatever... That is such an elementary fact, it is sometimes taken as the definition of prime. Your original question didn't refer to this at all anyway. Apr10 comment Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2$.Of course they are ALL possible divisors, this is easy to prove. Suppose you have your unique prime factorization of say $b$. And suppose another number $a$ divides $b$. Then we can apply the FTA to $a$ and get another prime factorization for $a$. If $p$ is a prime in the factorization of $a$, it must be in the factorization of $b$. Why? Well, you can use the property that if a prime number divides a product, it must divide one of the factors. So you have proven that every divisor of $b$ has to be made out of the primes in the factorization of $b$. Done! Apr8 comment Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2$.What do you mean? Mathematics is not a subjective field. Read the Fundamental Theorem of Arithmetics carefully. It says that every number can be descomposed uniquely (up to order) into a product of primes. Your number is already in such a form. So if you take all combinations of the factors, you get ALL divisors. Apr8 answered Let $p_1 , p_2$ be prime. Then prove that the only divisors of $p_1 p_2$ are $1 , p_1 , p_2 , p_1 p_2$. Apr7 accepted Equivalence of measures and $L^1$ functions Apr6 comment Equivalence of measures and $L^1$ functionsSo for example in the case of $X=[0,1]$, the Lebesgue measure $m$ and the Gauss measure $\mu(B) = \frac{1}{\log 2}\int_B \frac{dx}{1+x}$ this is ok. But how do I prove it? Apr5 comment Equivalence of measures and $L^1$ functionsI see! And what about if the function $g$ is continuous. Your function has a jump at $0$.