For $0\leq x \leq 1$, let $$f_n(x) = \frac{n}{1+n}, \ \text{if x is irrational} \\ \hspace{15mm}0, \qquad \text{if x is rational}$$
Let $f(x) = \lim_{n \to \infty} f_n(x)$. Then to find out whether $f$ s riemmann integrable or not?
According to me $$f(x) = 1, \ \text{if x is irrational} \\ \hspace{15mm}0, \qquad \text{if x is rational}$$
Clearly $f$ is not riemann integrable. But it is given in my notes $f$ is riemann integrable. Where i am mistaken?
Take any partition of $[0,1]$, say $0=x_0<x_1<x_2<\cdots<x_n=1$, then the difference between the upper and lower sum is $$U(f,(x_i))-L(f,(x_i))=\sum_{i=0}^{n-1}(x_{i+1}-x_i)\left(\sup_{x\in[x_i,x_{i+1}]}f(x)-\inf_{x\in[x_i,x_{i+1}]}f(x)\right)$$ Now for any partition $\sup_{x\in[x_i,x_{i+1}]}f(x)=1$ and $\inf_{x\in[x_i,x_{i+1}]}f(x)=0$. Hence for any partition $$U(f,(x_i))-L(f,(x_i))=\sum_{i=0}^{n-1}(x_{i+1}-x_i)=1$$ So $f$ is not Reimann integrable.