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.


1 Answer 1


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?

  • $\begingroup$ 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. $\endgroup$
    – Balbadak
    Nov 14, 2018 at 6:55
  • $\begingroup$ I opened a spearate question for the above comment. $\endgroup$
    – Balbadak
    Nov 14, 2018 at 7:53
  • $\begingroup$ OFC not… take $$\tilde{f_n} = 1_{[0,1]} + f_n$$ then $$ \tilde{f} = 1_{[0,1]}$$ and so your condition holds but still $$\int \tilde{f_n} = 2 \not\to 1 = \int \tilde{f}$$ $\endgroup$
    – Gono
    Nov 14, 2018 at 11:50

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .