Showing that $ \lim_{n\to\infty}\int_{0}^{1}f_n dx = \int_{0}^{1}f dx$ where $(f_n)$ are sequence of continuous functions converging to $f$ pointwise

Question

I am trying to figure out what is wrong with my proof for the below false statement:

Let $(f_n)$ be a sequence of continuous functions that converge pointwise to some Riemann-integrable function $f$. Then, for any $x\in[0,1]$,

$$\lim_{n\to\infty}\int_0^1 f_n(x) dx=\int_0^1 f(x) dx$$

My attempt at proof:

Let $x\in[0,1]$ and $\epsilon>0$. Since $f_n\to f$ pointwise, there is some $N\in\mathbb N$ s.t. for any $n>N$, $|f_n(x)-f(x)|<\epsilon$. Then,

\begin{align} \left|\int_0^1 f_n(x)dx - \int_0^1 f(x)dx\right| &= \left|\int_0^1 [f_n(x) - f(x)]dx\right| \\ &\le \int_0^1 |f_n(x) - f(x)|dx \\ &\le \epsilon \cdot (1-0)=\epsilon \end{align}

I think all my steps are correct. Did I overlook something here?

Of course! $x$ takes all the values of the interval of integration. — Anne Bauval, Feb 04 '24 at 23:27
@AnneBauval I thought $N$ depends on $n$. The expression $f_{N(x)}(x)$ doesn't make sense. — Kamal Saleh, Feb 04 '24 at 23:31
No, $N$ depends on $x$ (and of course on $\epsilon$): for every $x\in[0,1]$ and every $\epsilon>0$, there exist an $N$ such that etc. — Anne Bauval, Feb 04 '24 at 23:36
And the expression $f_{N(x)}(x)$ does make sense but is of no use. — Anne Bauval, Feb 04 '24 at 23:37
You might be thinking of the fact that continuity implies uniform continuity on a compact interval. It's not true that convergence implies uniform convergence even for a sequence of continuous functions on a compact interval. — anomaly, Feb 05 '24 at 00:01

score 3 · Accepted Answer · answered Feb 04 '24 at 23:38

The false bit is "Since $f_n\to f$ pointwise, there is some $N\in\mathbb N$ s.t. for any $n>N$, $|f_n(x)-f(x)|<\epsilon$".

What you wrote looks like pointwise convergence but you used it as if it were uniform convergence further down for lack of proper logical clarity.

Pointwise convergence is:

$$\forall x \exists N(x)\in\mathbb N / \forall n>N(x) |f_n(x)-f(x)|<\epsilon$$

But instead you used uniform convergence, which is

$$ \exists N \in\mathbb N / \forall x \forall n>N |f_n(x)-f(x)|<\epsilon$$

Obviously you can integrate only the second case, otherwise, as a comment states, your N depending on $x$, you never can be sure that it will not tend to infinity somewhere in the interval.

And as you know, uniform convergence of a continuous function on a compact allows interchanging the order of the limit and the integral.

score 3 · Answer 2 · answered Feb 04 '24 at 23:58

Counterexample:

Let $\{f_n\}_{n\in\mathbb N}$, where $f_n:[0,1]\rightarrow\mathbb R,$ with \begin{equation*} f_n(t) = \left\{\, \begin{array}{ccl} 0 & \quad \text{if}\quad & t\in \Big[ 0,\frac{1}{3n}\Big), \\ \quad 9n^{2} t-3n & \quad \text{if}\quad & t\in \Big[\frac{1}{3n},\frac{2}{3n}\Big), \\ \quad 9n-9n^2t & \quad \text{if}\quad & t\in \Big[\frac{2}{3n},\frac{3}{3n}\Big), \\ 0 & \quad \text{if}\quad & t\in \Big[ \frac{3}{3n},1\Big]. \end{array} \right. \end{equation*} Then $f_n(t)$, converges pointwise to 0, for every $x\in [0,1]$. However, for every $n \in \mathbb N$ \begin{equation*} \int_0^1 f_n(t)\,dt = 1 \qquad\text{while}\qquad \int_0^1 f(t)\,dt = 0. \end{equation*}

Hi there, can you please take a look at this question https://math.stackexchange.com/questions/4851938/revisited-question-lebesgue-measure-of-symmetric-difference-is-0? Thank you! — Hermi, Feb 05 '24 at 06:33

Showing that $ \lim_{n\to\infty}\int_{0}^{1}f_n dx = \int_{0}^{1}f dx$ where $(f_n)$ are sequence of continuous functions converging to $f$ pointwise

2 Answers2