Does $f_n \to f$ pointwisely imply $\int f_n \to \int f$ for conditionally Riemann integrable functions?

Let $$f_n:\mathbb{R}^n\to\mathbb{R}$$ be a sequence of conditionally (improper) Riemann integrable functions that pointwisely converges to $$f:\mathbb{R}^n\to\mathbb{R}$$ which is also conditionally Riemann integrable. Does $$\int f_n \to \int f$$? If they were absolutely Riemann integrable or Lebesgue integrable, it would be true by the dominated convergence theorem. I'd like to know it is also true (with some additional conditions) for conditionally converging integrals.

It's not true, even for absolutely Riemann integrable functions as you stated.

Take $$f_n = n1_{\left(0,\frac{1}{n}\right]}$$ then $$f_n \to f$$ pointwise with $$f \equiv 0$$

But $$\int f_n = 1 \not\to 0 = \int f$$

You cannot use dominated convergence theorem here although absolutely riemann integrability… why?

• How about $0<\int |f| <\infty$? Can this be a sufficient condition? Intuitively, this may prohibit accumulation of the mass of $f_n$ in an infinitely small region. Nov 14 '18 at 6:55
• I opened a spearate question for the above comment. Nov 14 '18 at 7:53
• OFC not… take $$\tilde{f_n} = 1_{[0,1]} + f_n$$ then $$\tilde{f} = 1_{[0,1]}$$ and so your condition holds but still $$\int \tilde{f_n} = 2 \not\to 1 = \int \tilde{f}$$
– Gono
Nov 14 '18 at 11:50