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How can I prove that $\int_0^1 \frac{1}{\sqrt{|x-1/2|}} \, dx < \infty.$ in Lebesgue measure?

My thought:

Using the Monotone Convergence theorem, but still I am stuck in applying it, could anyone help me please?

1 Answers1

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$$\int_0^{1/2}\frac{1}{\sqrt{\frac{1}{2}-x}}\,\mathrm d x\quad \text{and}\quad \int_{1/2}^1\frac{1}{\sqrt{x-\frac{1}{2}}}\,\mathrm d x$$ are convergent improper integrals of positives functions. So, obviously $x\mapsto \frac{1}{\sqrt{|\frac{1}{2}-x|}}$ is Lebesgue integrable on $(0,1)$.

Surb
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  • You should be able to compute them ;-) What is the derivative of $\sqrt{x-1/2}$ of $\sqrt{1/2-x}$ ? – Surb Mar 10 '20 at 14:14
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    No, with Lebesgue integral, you integrate on open sets. – Surb Mar 10 '20 at 14:19
  • To be more precise, the set ${0}$ and ${1}$ are null sets, so integrating with or without them is the same. – ViktorStein Mar 10 '20 at 15:08
  • but we do not calculate lebesgue integrals using derivatives. –  Mar 10 '20 at 15:17
  • @ViktorGlombik I wanted to use the theorem that if the function is bounded and remann integrable on a closed interval then it is Lebesgue integrable.... or what theorem is you referring to ? –  Mar 10 '20 at 15:22
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    @Mathstupid I am referring to $\int_{[0,1]} = \int_{(0,1)}$. – ViktorStein Mar 10 '20 at 15:42
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    if an function is Riemann integrable, then its Lebesgue integral is computed in the same way. – Surb Mar 10 '20 at 20:13