# Sequence of $f_n \in R[0, 1]$ that converges pointwise to $f \in R[0, 1]$ such that $\lim_{n \to \infty} \int_0^1 f_n dx \neq \int_0^1 f dx$.

Question:

Find a sequence of functions $f_n \in R[0, 1]$ that converges pointwise to $f \in R[0, 1]$ such that $\lim_{n \to \infty} \int_0^1 f_n dx \neq \int_0^1 f dx$. ($R$ means Riemann-Stieltjes integrable)

Attempt:

Pick \begin{align*} f_n(x) = \begin{cases} n \quad x \in (0,1/n) \\ 0 \quad else \end{cases}. \end{align*} Let $f(x)$ be the zero function. Hence, $\lim_{n \to \infty} f_n = f$. Then, we have $$1 = \lim_{n \to \infty} \int_0^1 f_n dx \neq \int_0^1 f dx= 0$$ and we're done.

Is this a suitable answer?

• What is the question? – copper.hat Nov 26 '14 at 3:10
• Very nice, indeed! :) – Swapnil Tripathi Nov 26 '14 at 3:12
• @copper.hat I was looking for verification of my attempt. I have modified my question and added a proof-verification tag. Sorry about that. – mathjacks Nov 26 '14 at 3:18

## 1 Answer

The answer is correct, provided you can justify that $f_n\in R[0,1]$. Maybe there's a theorem about step functions being integrable, or about piecewise continuous functions, or you just do it by hand for this example.

If you wanted to use the theorem that continuous functions on $[0,1]$ are Riemann integrable, a modified example would be useful: "skinny triangles": $$f(x)=\begin{cases} n^2x,\quad &0\le x\le 1/(2n), \\ n-n^2x, \quad & 1/(2n)\le x\le 1/n; \\ 0,\quad &1/n\le x\le 1 \end{cases}$$