# Uniform convergence of integrable functions

1. Apr 10, 2006

### hooker27

This question arised in my last math class:

If a sequence of functions $$f_n$$ uniformly converges to some $$f$$ on $$(a, b)$$ (bounded) and all $$f_n$$ are integrable on $$(a, b)$$, does this imply that $$f$$ is also integrable on $$(a, b)$$ ??
($$f_n$$ do not necessarily have to be continous, if they were, the answer would be obvious)

Note: It is not certain, which type of integral is meant, it can be Newton, Riemann or Lebesgue. Let me please know if the answer depends on which type of integral is used.

- If it is true, could you please tell me where (on www) I might find a proof??
- If it is not true, could you please show me a sequence for which it is not true??

Thanks a lot, H.

2. Apr 10, 2006

### AKG

An easy way to see Riemann integrability is to note that if fn is continuous at x for all n, then f is continuous at x. Therefore, a discontinuity point of f must be a discontinuity point of some fn (the converse need not be true). Thus, the set of discontinuity points of f is contained in the union of the discontinuity points of the fn. For each fn, the set of discontinuity points is a zero set (assuming Riemann integrability), so the set of discontinuity points of f is contained in the countable union of zero sets, hence is contained in a zero set, hence is a zero set, so f is Riemann integrable.

Lebesgue integrability is preserved as well, I believe. Try looking up the Lebesgue dominated convergence theorem.

3. Apr 11, 2006

### HallsofIvy

What, precisely, do you mean by "Newton" integrable? I don't recognise that term.

4. Apr 11, 2006

### hooker27

Maybe there is s different name for that in english. What I know as Newton's integral is this:
if a function f is defined on (a,b) and there exists some F such that $$F'(x) = f(x)$$ for all x from (a,b) (in other words - the function F is an antiderivative of f on (a,b) ) then Newton's integral of f over (a,b), denoted as
$$\int_{a}^{b}f(x)dx$$
is defined as
$$\lim_{x\to b-}F(x)-\lim_{x\to a+}F(x)$$.

So what I call "Newton integrable" is:
- function must have an antiderivative
- the limits (above) of the antiderivative must exist
- the expression with the limits (above) must be well defined (not $$\infty -\infty$$ etc.)

As for my previous question: of all discontinous functions $$f_n$$ have antiderivatives, does their uniform limit $$f$$ also have an antiderivative?