# Tag Info

4

The easiest way is to notice that $f$ differs from the zero function (which is measurable) on a null-set. As null-sets are always measurable, $f$ is measurable. You can also do it explicitly: for $2^n\leq a$, $f^{-1}(a,\infty)=\{n,n+1,n+2,\ldots\}$. for $a=0$, $f^{-1}(a,\infty)=\mathbb Z$. for $a<0$, $f^{-1}(a,\infty)=\mathbb R$.

4

You have that $$\int_{[0,1]}fd\lambda=\int_{[0,0.1]}fd\lambda+\dots+\int_{[0.9,1]}fd\lambda.$$ Note that $f(x+0.1)=f(x)$ to get that $$\int_{[0,1]}fd\lambda=10\times\int_{[0,0.1]}fd\lambda.$$ Now write $$\int_{[0,0.1]}fd\lambda=\int_{[0,0.01]}fd\lambda+\dots+\int_{[0.09,0.1]}fd\lambda=0.01\times[1+\dots+9]$$ to finally get $$\int_{[0,1]}fd\lambda=10\times ... 3 Assume that f is continuous. Let \epsilon > 0 and consider$$ G_k := \{x \in [0,1] : |f_k(x) - f(x)| \leq \epsilon/2\} $$and set$$ F_n = \cap_{k=n}^{\infty} G_k = \{x \in [0,1] : \sup_{k\geq n}|f_k(x) - f(x)| \leq \epsilon/2 \} $$Then F_n is closed and by pointwise convergence$$ [0,1] = \bigcup_{n=1}^{\infty} F_n $$So by Baire Category, ... 3 If we allow \mu to be an infinite measure, then at least precisely the same formulation will not work. Let \mu = \infty \cdot \lambda be the Lebesgue measure "multiplied by \infty", i.e. \mu(E) = 0 if \lambda(E) = 0 and \mu(E) = \infty otherwise. Then L^\infty (\mu) = L^\infty (\lambda) is infinite dimensional, although \mu only assumes two ... 3 The sequence (a_n)_n is Cauchy: If m>n>N then$$|a_m-a_n|\le\sum_{k=n}^{m-1}|a_{k+1}-a_k|\le \sum_{k=N}^\infty|a_k-a_{k+1}|$$and the latter is <\epsilon for N large enough for this expresses exactly that \sum_{k=1}^\infty|a_k-a_{k+1}|  converges. Hence with a:=\lim_{k\to\infty} a_k, we have \lim_{k\to\infty}|a_1-a_k|=|a_1-a|. 2$$|(a_1-a_r)-(a_1-a_s)|=|a_r-a_s|= |\sum_{k=r}^{s-1} a_{k+1}-a_k | \le \sum_{k=r}^{s-1} |a_{k+1}-a_k |$$This is convergent. 2 \cos x = 1 -\frac12x^2+\frac{1}{24}x^4+\cdots e^{-\frac12x^2} = 1 -\frac12x^2+\frac{1}{8}x^4+\cdots So if \varepsilon is small and |x| < \varepsilon, they differ by at most \approx \frac{1}{12}\varepsilon^4. 2 Assume that \lim_{n \to \infty} x_{n+m} = x. Then, for every \varepsilon > 0, there exists N_0 such that if n > N_0 then |x_{n+m} - x| < \varepsilon. If we define N := N_0 + m then we see that if n > N then |x_n - x| < \varepsilon. 1 The f_n are continuous functions on \mathbb{R}, but the poinwise limit of f_n(x) is not, since the uniform convergence preserve the continuity you can conclude that f_n can't converge uniformly on all \mathbb{R}. 1 The integral bound$$\int_0^t |x(s)| ds \le t^4$$don't say much when t is large. In particular, the set$$L = \{ f\in C[0,1] : f(x) = 0 \text{ on } [0,1/2], \ \|f\|_\infty \le 1\}$$is a subset of K. This L is as big as \{ f\in C[2/3,1]: \|f\|_\infty \le 1\}, so L cannot be compact and so is K. 1 If I_{1} and I_{2} are sets (such as intervals of real numbers), define I_{1} + I_{2} to be their disjoint union and I_{1} \times I_{2} to be their Cartesian product. Since the Cartesian product distributes over disjoint unions,$$ (I_{1} + I_{2}) \times (I_{1} + I_{2}) = (I_{1} \times I_{1}) + (I_{1} \times I_{2}) + (I_{2} \times I_{1}) + (I_{2} ...

1

Assuming we're dealing with the Borel $\sigma$-algebra on $\mathbb{R}$. You should how that $\{[q_n,q_m): q_n,q_m \in Q\}$ generate the Borel $\sigma$-algebra. This is simple: Let $a_n$ be a decreasing sequence of rational numbers which tends to $-\infty$ , and $b_n$ be an increasing sequence of rational numbers which converges to $b$, and $b\in ... 1 The nontrivial part is to prove that the inverse image condition is sufficient. I am assuming that measurability of a function is defined with respect to the Borel sigma-algebra on the reals (for the larger Lebesgue sigma-algebra the statement is false). The Borel sigma-algebra is generated by open intervals so it is enough to prove that the inverse image of ... 1 Notice that:$|\sum_{k=1}^{n} a_k-a_{k+1} | = |a_n - a_1 | \leq \sum_{k=1}^{n} |a_k-a_{k+1}|$for each$n \in \mathbb{N}1 It's true. I'll show transitivity. Define the relation between sequences of closed intervals as: \begin{align} ([a_n,b_n])_{n\in\Bbb N} \simeq ([c_n,d_n])_{n\in\Bbb N} \iff & \forall n\in\Bbb N: \\ &\quad a_n \le d_n, \text{ and }\tag{i} \\ &\quad c_n\le b_n.\tag{ii} \end{align} A sequence of nested intervals is a sequence with the properties ... 1 Given\epsilon > 0$and$x$, we want to show that there exists a$\delta>0$such that$|F(x)-F(y)|<\epsilon$whenever$|x-y|<\delta$. Since$[-x,x]$is compact, the minimum is attained at some point(s)$t_0\in[-x,x]$. Suppose first that$\pm x$are not points of minimum, i.e. the minimum is attained strictly inside. Let$\delta$be such that ... 1$\ell^1$is a metric space and its closure equals to the set of all elements that have distance$0$from$A$. On one side, any element$g$of $$B:=\{f\in\ell^1:\,\,\forall\,\,n\,\,|f(n)|\leq \frac{1}{2^n}\}$$ has distance$0$from$A$, because it either in$A$or a limit of elements in$A$. On the other side,$\ell^1\setminus B\$ is open (being equal to ...

1

The uniform norm is bounded by the D norm but not vice versa. Consider a sequence of sine waves scaled so that the frequency increases very rapidly and the amplitude converges slowly to 0.

Only top voted, non community-wiki answers of a minimum length are eligible