Let f(x) be defined on [0, 1] as follows:\[ f(x) = 1\begin{cases}\text{1 when } x \text{ is rational}\\\text{-1 when } x\text{ is irrational}\end{cases}\]
Correct Answer: Description for Correct answer:
Let \(P=\{ 0=x_{0},x_{1},x_{2},...,x_{n}=1 \}\)
be a partition.
\(I_{r}=\left[ x_{r-1},\ x_{r} \right]\)
\(r=0,1,2,...,\)
\(\delta_{r}=x_{r}-x_{x-1}\)
\(M_{r}=\underset{x\in I_{r}}{lub}f(x)=1\)
\(m_{r}=\underset{x\in I_{r}}{glb}f(x)=1\)
[Because in any interval \(I_{r}\), there exists rationals and irrationals]
Now,
\(U(P,f)=\sum\limits_{r=1}^{n}M_{r}\delta_{r}\)
\(=\sum\limits_{r=1}^{n}1(x_{r}-x_{r-1})\)
\(=1[(x_{1}-x_{0})+(x_{2}-x_{1})+...(x_{n}-x_{n-1})]\)
\(=x_{n}-x_{0}=1-0=1\)
\(L(P,f)=\sum\limits_{r=1}^{n}m_{r}\delta_{r}\)
\(=\sum\limits_{r=1}^{n}(-1)(x_{r}-x_{r-1})\)
\(=-1[(x_{1}-x_{0})+(x_{2}-x_{1})+...+(x_{1}-x_{n-1})]\)
\(=-1[x_{n}-x_{0}]\)
\(=-1[1-0]=-1\)
\(\therefore\int\limits_{0}^{\overline{1}}=\lim\limits_{n\rightarrow \infty}U(P,f)=1\)
\(\int\limits_{\overline{0}}^{1}f=\lim\limits_{n\rightarrow \infty}L(P,f)=-1\)
Thus,
\(\int\limits_{\overline{0}}^{1}f\ne \int\limits_{0}^{\overline{1}}\)
\((i.e.)\ f\notin R[0,1]\)
Therefore \(f\) is not Riemann integrable.
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