Is there integrable function sequence which is uniformly converges to not integrable function?

Solution 1:

This is not possible; if $f_n$ is a sequence of Riemann-integrable functions that converge uniformly to $f$ , then $f$ is Riemann-integrable, and $$\int \lim_{n\rightarrow\infty}f_n= \int f$$

Notice that , for an enumeration {$q_1,q_2,...$} of $\mathbb Q$ , the sequence {$f_n$} given by $f_n(x)=1$ if $x=q_n$ and 0 otherwise, converges to $\chi_{\mathbb Q}$, which is not Riemann-integrable , e.g., by an Upper-bound is 1 and lower-bound is 0 argument, but theconvergence is not uniform.

For a proof of the Riemann integrability of $f$, we use upper- and lower- sums for the Riemann sum for $f$ and for $f_n$ , the fact that the $f_n$ are Riemann-integrable, and that $|f_n(x)-f(x)|\rightarrow 0$ uniformly.

1) We have by uniform convergence, for $n=1,2,...$ , that $1/n-f_n(x)<f(x)<1/n+f_n(x)$

2)Since $f_n$ is Riemann integrable, $ U(f_n, ||P_n||)-L(f_n, ||P_n||)<1/m $ , where U,L, are the upper- and lower sums, and ||P_n|| is the partition width for the sums.

3)We need to show that from 1), 2), it follows that there is a partition withdth ||P|| so that$U(f,||P||)-L(f.||P||)<1/n$

4)Once we have showed that $f$ is Riemann-integrable, we have : $\int f_n$ - $\int f= \int(f_n-f)=0$ , so $\int f_n \rightarrow \int f$ Can you take it from here?