3 votes
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Show that $\int |g_n - g|\to 0$ if and only if $\int |f_n| \to \int |f|$

Suppose $\int |g_n| \to \int |g|$. Since $|g_n - g| \le |g_n| + |g|$ for all $n$ and $g_n \to g$ a.e., Fatou's lemma yields $$\varliminf_{n\to \infty} \int (|g_n| + |g| - |g_n - g|) \ge \int 2|g|$$ ...
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  • 40.2k
3 votes
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Double integral over the triangle 0<y<x<1

We have that: $\{(x,y)\in\Bbb R^2:0 < y< x < 1\}=\{(x+y, y)\in\Bbb R^2: 0<y<1, 0<x<1-y\}$ So your integral is $$\int_0^1 \sqrt{1-y~}\int_0^{1-y}\dfrac 1{\surd x}\,\mathrm d x\,\...
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2 votes

Why dont we only consider complete measure spaces?

Here are two arguments. I don't think you are addressing them (maybe because I don't speak categories and don't know much about Bochner integration). I use $(\Omega, \mathcal{A}, \mu)$ for a ...
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1 vote
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Show that $t\mapsto\int_0^tf(s)\:{\rm d}s$ is absolutely continuous

If you are talking about absolute continuity on the whole of $[0,\infty)$ (in the sense for every $\epsilon >0$ there exists $\delta >0$ such that $\sum |f(b_i)-f(a_i)| <\epsilon $ whenever $(...
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  • 2,186
1 vote
Accepted

Does $\|f_n-f\|_1 \to 0$ where $f_n$ is the mean of $f$ over a uniform partition in $n$-bins of the domain?

Hints: If $f$ is continuous then $f_n (t) \to f(t)$ for each $t$ and $(f_n)$ is uniformly integrable. This implies that $f_n \to f$ in $L^{1}$. For the general case choose a continuous function $g$ ...
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  • 2,186
1 vote
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A diagonal differentiation theorem

For $f \in L^2([0,1])$, let $$T_n(f):= n\cdot\sum_{k=1}^n\int_{I_{n,k}\times I_{n,k}}|f(s)-f(t)|\, ds\, dt \,.$$ As noted in the problem, applying Cauchy-Schwarz twice gives $$\frac{T_n(f)^2}{n^2} \le ...
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  • 12.9k
1 vote
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If $g$ is bounded, nondecreasing and right-continuous, how do we show $\int f(t)\:{\rm d}g(t)=\int f(t)g'(t)\:{\rm d}t$?

If $g$ is the Cantor function then $g'=0$ almost evrywhere so the right hand side is $0$ for any $f$. Take $f=\chi_{[0,1]}$ to get a contradiction.
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  • 2,186
1 vote
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limits of 2 lebesgue integrals

For 1) use the fact that $|\sin t| \leq t$ for all $t >0$ so a dominating function is $\frac 1 {1+y^{2}}$. For 2) use the fact that $(1+y^{2})^{m} \geq 1+my^{2}$ so a dominating function is the ...
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  • 2,186

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