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Suppose $f$ is integrable on $[0,b]$. Let $g(x)=\displaystyle\int^b_x\frac{f(t)}t\;dt,0

Now I want to prove that $g$ is integrable on $[0,b]$. But I don't know how to show that.

Can anyone give me some hints?

Thank you very much.

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    Did you mean integrable on $[0,b]$?2011-12-08
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    I am so sorry for that. I just want to say that $f$ is integrable.2011-12-08
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    First $g$ is continuous on $(0,b]$. Are you talking about the Riemann or Lebesgue integrals? For Lebesgue all that remains is to show $\int_0^b |g(x)|\,dx < \infty$.2011-12-08
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    I mean Lebesgue integrals.2011-12-08
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    @GEdgar How to show that $g$ is continuous? Can you give me more details? Thank you .2011-12-08
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    First show $g$ is continuous on $(0,b]$. Let $u \in (0,b]$, we must show $g$ is continuous at $u$. I will do the case $u$c,d$ with $0$f(t)/t$ is integrable on $[c,d]$. Therefore $g$ is continuous at $u$. – 2011-12-08
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    That's right.I understand.2011-12-08
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    And thank you very much.2011-12-08
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    Hint: apply Fubini's theorem.2011-12-08
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    If you now understand how to solve the problem, you should write up an answer. It may seem odd to answer your own question, but site policy actually encourages doing that. Then you can accept your answer (and answers to some of your other questions, too!).2011-12-09

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Fix $\delta>0$. Then we have, using Fubini's theorem (as @Zarrax suggests) \begin{align*} \int_{\delta}^b|g(x)|dx&\leq\int_{\delta}^b\int_x^b\left|\frac{f(t)}t\right|dtdx\\ &=\int_{{\delta\leq x\leq t\leq b}}\frac{|f(t)|}tdtdx\\ &=\int_{\delta}^b\int_{\delta}^t\frac{|f(t)|}tdxdt\\ &=\int_{\delta}^b\frac{t-\delta}t|f(t)|dt\\ &\leq \int_{\delta}^b|f(t)|dt\\ &\leq \int_0^b|f(t)|dt,\\ \end{align*} and we are done since the map $\delta\mapsto \int_{\delta}^b|g(x)|dx$ is decreasing.