# Sequence of integrable functions (f_n) conv. to f

Homework Statement:
Let ##(f_n)## be a sequence of integrable functions on ##[a,b]##, and suppose ##f_n \rightarrow f## uniformly on ##[a,b]##. Prove that ##f## is integrable and
$$\int_a^b f = \lim_{n\rightarrow\infty} \int_a^b f_n$$
Relevant Equations:
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##\textbf{Attempt at solution}##: If I can show that ##f## is integrable on ##[a,b]##, then for the second part I get :

Let ##\frac{\varepsilon}{b-a} > 0##. By definition of uniform convergence, there exists ##N = N(\varepsilon) > 0## such that for all ##x \in [a,b]## we have ##\vert f(x) - f_n(x) \vert < \frac{\varepsilon}{b-a}##. This gives us,

$$\vert \int_a^b f(x)dx - \int_a^b f_n(x)dx \vert = \vert \int_a^b f(x) - f_n(x) dx \vert \le \int_a^b \vert f(x) - f_n(x) \vert dx < \int_a^b \frac{\varepsilon}{b-a}dx = \varepsilon$$ when ##n > N##.

It follows ##\lim_{n\rightarrow\infty} \int_a^b f_n(x) dx = \int_a^b f(x) dx## for all ##x \in [a,b]##. []

For the first part I am stuck. Let ##\varepsilon > 0##. Then I need a partition ##P## of ##[a,b]## such that ##U(f, P) - L(f, P) < \varepsilon##. We're given that for any ##n##, there is a partition ##P## such that ##U(f_n, P) - L(f_n, P) < \varepsilon##. How can I proceed?

Delta2

PeroK
Homework Helper
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For the first part I am stuck. Let ##\varepsilon > 0##. Then I need a partition ##P## of ##[a,b]## such that ##U(f, P) - L(f, P) < \varepsilon##. We're given that for any ##n##, there is a partition ##P## such that ##U(f_n, P) - L(f_n, P) < \varepsilon##. How can I proceed?

The upper and lower sums work like integrals in a way, so you can use a similar technique for these as you used to show part b).

fishturtle1
member 587159
Here is a measure theoretic solution which you may appreciate later (maybe you treat this theorem in a first course, probably not).

We use the following result (see any text on measure theory or more advanced texts on Riemann-integration).
A function is Riemann-integrable if and only if it is bounded and its set of discontinuities has (Lebesgue) measure ##0##.

The points where atleast one ##f_n## is discontinuous has measure zero so the limit of the ##f_n's## is almost surely continuous as uniform convergence preserves continuity. Hence the limit is integrable.

Last edited by a moderator:
fishturtle1
The upper and lower sums work like integrals in a way, so you can use a similar technique for these as you used to show part b).
We'll show ##f## is integrable on ##[a,b]##.

Proof: Let ##\frac{\varepsilon}{4(b-a)} > 0##. Then there is ##N = N(\frac{\varepsilon}{4(b-a)}) > 0## such that for ##x \in [a,b]## and ##n > N## implies $$\vert f_n(x) - f(x) \vert < \frac{\varepsilon}{4(b-a)}$$ i.e
$$-\frac{\varepsilon}{4(b-a)} < f(x) - f_n(x) < \frac{\varepsilon}{4(b-a)}$$
Fix ##n > N##. For ##\frac{\varepsilon}{2} > 0##, there exists partition ##P = \lbrace a = t_0 < t_1 < \dots < t_k = b \rbrace## such that ##U(f_n, P) - L(f_n, P) < \frac{\varepsilon}{2}##.

Now, $$U(f - f_n, P) = \sum_{i=1}^{k} M(f - f_n, [t_{i-1}, t_i]) \cdot (t_i - t_{i-1}) < \sum_{i=1}^{k} \frac{\varepsilon}{4(b-a)} \cdot (t_i - t_{i-1}) = \frac{\varepsilon}{4(b-a)} \sum_{i=1}^{k} (t_i - t_{i-1}) = \frac{\varepsilon}{4}$$

Also, $$L(f - f_n, P) = \sum_{i=1}^{k}m(f - f_n, [t_{i-1}, t_i])\cdot (t_i - t_{-1}) > \sum_{i-1}^{k} -\frac{\varepsilon}{4(b-a)}\cdot(t_i - t_{i-1}) = - \frac{\varepsilon}{4}$$

Thus, $$U(f, P) - L(f, P) \le U(f - f_n, P) + U(f_n, P) - (L(f - f_n, P) + L(f_n, P)) < \varepsilon$$
It follows that ##f## is integrable on ##[a,b]##. []

Last edited:
Delta2
Here is a measure theoretic solution which you may appreciate later (maybe you treat this theorem in a first course, probably not).

We use the following result (see any text on measure theory or more advanced texts on Riemann-integration.

The points where atleast one ##f_n## is discontinuous has measure zero so the limit of the ##f_n's## is almost surely continuous as uniform convergence preserves continuity. Hence the limit is integrable.