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Let the unbounded function $f(x)$ on the interval $[0,1]$ be defined as

$$f(x)=\left\{\begin{array}{l l}\frac{1}{x}&x\text{ in }(0,1]\\ f(x)=0&x=0\end{array}\right.$$

Show that $f(x)$ is not Riemann integrable.(Hint:infinity is not a real number.)

joriki
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Iclal Demir
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  • Let the unbounded function (f(x)) on the interval ([0,1]) be defined as (\displaystyle f(x) = \left{\begin{array}{cr}\frac{1}{x} & x\in(0,1]\0 & x =0\end{array}\right.). Show that (f(x)) is not Riemann integrable. (Hint: infinity is not a real number) – Iclal Demir Mar 12 '13 at 21:09
  • You can estimate the integral on $[\delta,1]$ with a simple Riemann sum. What happens for $\delta \to 0$? – Quickbeam2k1 Mar 12 '13 at 21:11
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    Alex asked you what have you tried, you repeat your exercise problem. That is just unacceptable. – mez Mar 12 '13 at 21:24
  • I'm sorry.It's the first time I have asked a problem here.I couldn't understand what should I say exactly.I tried to write the definition of Riemann Integration and with the help of your comment I'm tring to use Darboux Integrable. – Iclal Demir Mar 12 '13 at 21:37

3 Answers3

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Apparently you want a Riemann sum argument. So take $n\geq 1$ and consider the partition $\mathcal P_n=\{1/n, 2/n,\ldots,(n-1)/n,1\}$ of $[0,1]$. The corresponding Riemann sum of $f$ is $$ R(f,\mathcal P_n)=\frac{1}{n}\sum_{k=1}^n\frac{1}{\frac{k}{n}}=\sum_{k=1}^n\frac{1}{k}. $$ Now you recognize the partial sum of the harmonic series. It can be proved without integration that this tends to $+\infty$ by grouping terms as follows $$ 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\ldots+\frac{1}{8}+\ldots $$ $$ \geq 1+ \frac{1}{2}+\left( \frac{1}{4}+ \frac{1}{4}\right)+ \left( \frac{1}{8}+ \frac{1}{8}+ \frac{1}{8}+\frac{1}{8}\right)+\ldots=1+\frac{1}{2}+\frac{1}{2}+\ldots $$ So these Riemann sums tend to $+\infty$ and therefore $f$ is not Riemann-integrable.

Julien
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First, what are your attempts on the problem?

A hint:

Recall that there is a theorem which said Riemann integrable and Darboux integrable are equivalent. Try a proof by contradiction, using the epsilon-partition definition of how it means to be "Darboux integrable". We should get a contradiction to the definition with a "certain" epsilon.

Cecile
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Well the antiderivative of $\frac{1}{x}$ is $\log(x)$ no show what happens when $x\to 0$

$$\int_0^1 f(x)\, \mathrm{d}x= \int_0^\varepsilon f(x) \, \mathrm{d}x + \int_\varepsilon^1 f(x)\, \mathrm{d} x$$ Hence $$ \int_0^1 f(x)\, \mathrm{d}x= \int_0^\varepsilon f(x)\, \mathrm{d}x+ \ln(1) - \ln(\varepsilon)=\int_0^\varepsilon f(x) \, \mathrm{d}x - \ln(\varepsilon)$$

Dominic Michaelis
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