9 votes
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Can we conclude that $f=g$ a.e. if $\int _Efd\mu =\int _Egd\mu $ for all measurable sets $E$?

There are measure spaces where all measurable sets have measure $0$ or $\infty$. In such a space, let $f$ and $g$ be two different positive constants. [EDIT] More generally, suppose there is a ...
Robert Israel's user avatar
4 votes
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Existence of a sequence $\mu(I_j \cap I_k)= 0$ such that $\lim_{j\to \infty}\mu(I_j) = b-a$

$\mu(I_j \cap I_k)= 0$ for $j \ne k$ implies that $\sum_j \mu(I_j)=\mu(\bigcup_j I_j) <\infty$ and hence, $\mu (I_j) \to 0$.
geetha290krm's user avatar
  • 37.3k
4 votes
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Doubts in calculatating $\int_0^\infty \frac{n\sin x}{1+n^2x^2}\,\mathrm{d}x$

Yes, your estimate is correct, as $$0\leq(nx-1)^2=n^2x^2-2nx+1$$ implies that $$1+n^2x^2\geq2nx,$$ and the rest of the estimates are clear. Now the reason why you can ignore the point $x=0$ is because ...
Lorago's user avatar
  • 9,639
2 votes
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Show that $\int_{\mathbb R^d}fd\mu=\sum_{x\in C}f(x)$ for countable $C\subset R^d$

For Case 3. We can proceed as follows. Let $f$ be a non-negative, measurable function. Set $$ f_n(x) = \sum_{k = 1}^n f(c_k) \cdot 1\{x = c_k\} = \begin{cases} f(c_k), \text{ if } \exists k \in \{...
Thành Nguyễn's user avatar
2 votes
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Is there a need to specify $g$ as non-negative for $\int gd\nu=\int gfd\mu$ to hold?

Normally, if $g$ is non-negative and measurable (rather than integrable) then one allows $\int g d\nu = +\infty$, where this means that the set $$ \{\int sd\nu: s \text{ a simple function}\}\subseteq \...
krm2233's user avatar
  • 4,583
1 vote
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Show that $\lim_{x\rightarrow a}F(x)$ exists finitely for $\int_a^b|F'(x)|dx<\infty$, $F\in C^1$

Suppose $y \in (a,b)$ and choose $y' \in (a,y)$. Then $F(y) = F(Y') + \int_{y'}^y F'(t)dt = \int 1_{[y',y]}(t) F'(t)dt$ Let $y_n \to b$ (with $y_n \in (a,b)$, of course), then since $|1_{[y',y_n]}(t) ...
copper.hat's user avatar
  • 173k
1 vote

Analysis of an expression involving a function on $\mathbb R^n$. Related to limits, supremums and translations.

About your approach. The function $f$ is bounded outside $B(0,1/2)$, i.e., $|f(y)| < C$ for some $C > 0$ and all $y \not \in B(0,1/2)$. Note that for every small enough $\xi$, the symmetric ...
Jordan Payette's user avatar

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