Sequence of integrable functions (f_n) conv. to f

[fishturtle1]

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?

 

[PeroK]

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).

 

[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.

 

[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 $[a,b]$. []

 

[fishturtle1]

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.

Thanks for posting this. I think I am learning about this, this upcoming semester.

 

[member 587159]

Thanks for posting this. I think I am learning about this, this upcoming semester.

It's a very powerful tool and actually weird it doesn't occur in much places. For example, it immediately implies that continuous functions, monotonic functions, step-functions, functions with countably many discontinuities (e.g. https://en.wikipedia.org/wiki/Thomae's_function) are all Riemann-integrable. It can also be used to quickly show that something is not Riemann-integrable. For example, an indicator function on the rationals or the Cantor set can not be Riemann-integrable (note that these become integrable if one considers the Lebesgue integral instead of the Riemann-integral).