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Let $m(t)$, $f(t)$ are functions on $[0;1]$ which are assumed to be measurable, $m(t)\neq0$ almost everywhere and $f(t)\geq0$ for all $t$ belongs to the interval $[0;1]$. Let me recall that $\text{sgn}m(t)=\begin{cases}0 \quad\text{if $m(t)=0$}&\\ 1\quad\text{if $m(t)>0$ and}&\\ -1\quad\text{if $m(t)<0$.}\end{cases}$.

It seems that there exist such functions $m$ and $f$ so that $\left|\int\limits_0^1 \left(\text{sign $m(t)$}\right)\cdot\,f(t)dt\right|<\infty$ but $\int\limits_0^1f(t)dt=\infty$, but I get stuck to construct such functions. So do there exist such functions?

Hai Minh
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1 Answers1

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Let $m(t)$ be $\sin(t)$ , and $f(t)= 1/t$. Now $\int\limits_0^1f(t)dt=\infty$ , and $\left|\int\limits_0^1\sin(t)f(t)dt \right|< \infty$. if you do not need $m(t)$ to be continuous , you can take $m(t) = (-1)^n$ for each $n \in {N}$.

Eli Elizirov
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  • I agree that your example is true in some sense. Here I assume that $f$ is determine on $[0;1]$, it means $f(0)$ is a real number. Unlickily, your example $f(0)=\infty$. – Hai Minh Jul 21 '13 at 08:56
  • ok than you can take $f(x)=1/(x-1/2)$ and if f(x) is a real number for any x in [0,1] than the integral of f wil be smaller than infinity – Eli Elizirov Jul 21 '13 at 09:29
  • You mean $f(x)=\frac1{x-1/2}$ if $x\neq 1/2$ and $c$ if $x=1/2$, where $c\in\mathbb R$ ? What is $m(x)$? – Hai Minh Jul 21 '13 at 09:39
  • $m(x)$ will be $sin(x)$ – Eli Elizirov Jul 21 '13 at 10:06