How do I show $\lim_{n \rightarrow \infty} \int_{0}^{1}f_{n}(x)dx=\int_{0}^{1}f(x)dx$.

Question

Let $f_{n}(x)$ and $f(x)$ be continuous functions on $[0,1]$ such that $\lim_{n \rightarrow \infty} f_{n}(x)=f(x)$ for all $x \in [0,1]$.

a)Can we conclude that $\lim_{n \rightarrow \infty} \int_{0}^{1}f_{n}(x)dx=\int_{0}^{1}f(x)dx$.

b) If $|f_{n}(x)|\leq 2017$ for all n and for all $x\in [0,1]$, can we conclude that $\lim_{n \rightarrow \infty} \int_{0}^{1}f_{n}(x)dx=\int_{0}^{1}f(x)dx$.

My attempt : For part a), my answer is No. I choose an example $f_{n}(x)=n^{2}x(1-x^2)^n$.

For part b), if I show that $f_{n}(x)$ converges uniformly then the given result is true.

For uniform convergence, Can I say if $|f_{n}(x)|$ is uniformly bounded and convergence implies uniform convergence.

Can anyone suggest some strong justification for part b)?

Answer 1

While uniform convergence would be sufficient for (b), it is not necessary with the given assumptions. For example, $f_n(x)=x^n$ is uniformly bounded for $x \in [0,1]$, but does not converge uniformly.*

But for uniformly bounded $f_n$, you can argue by way of the Dominated Convergence Theorem.

*Indeed, for $0 \leq x < 1$ and $0 < \epsilon < 1, |f_n(x)-f(x)|=|x^n|<\epsilon$ if and only if $x < \epsilon ^ {\frac{1}{n}}$. Since $\epsilon^{\frac{1}{n}} < 1$ for all $n$, no $N$ works of all $x$ sufficiently close to $1$, for $N$ in the definition of uniform convergence.

Answer 2

Your example for the a part is fine.

For b, the answer slightly depends on the "meaning" of $\int_0^1$, i.e. what kind of integral it is. If it's Lebesgue integral, then The Dominated Convergence Theorem suggested by gfppoy will do the job. But if it's Riemann-integral, it's not really helpful. But there exists a (not as famous) version of the DCT for the Riemann-integrals as well, called Arzelà's Dominated Convergence Theorem, which states the following:

"Let ${f_n}$ be a sequence of Riemann-integrable functions defined on a bounded and closed interval $[a, b]$, which converges on $[a, b]$ to a Riemann-integrable function $f$. If there exists a constant $M>0$ satisfying $|f_n(x) | < M$ for all $x \in [a, b]$ and for all $n$, then $$\lim_{n \to +\infty} \int_a^b |f_n(x)-f(x) | \mathrm{d}x=0.$$ In particular, $$\lim_{n\to +\infty}\int_a^b f_n(x)\mathrm{d}x = \int_a^b\lim_{n\to +\infty} f_n(x)\mathrm{d}x=\int_a^bf(x)\mathrm{d}x."$$

Answer 3

Part b):Arzela Ascoli guarantees that there exists a uniformly convergent subsequence, and the pointwise convergence of $ f_n $ to $ f $ says that this uniformly convergent subsequence must converge to $ f $ so we have that there is a sequence $k_n$ that $ f_{k_n} ->f$ uniformly .No we use contradiction:Suppose then that $ f_n $ does not converge uniformly to $ f $. Then, there exists an $ \epsilon > 0 $ and a sequence $ x_{k_m} $ in $ [0, 1] $ such that $ |f_{k_m}(x_{k_m}) - f(x_{k_m})| \geq \epsilon $. However, this tells us that no subsequence of $ f_{k_n} $ converges uniformly to $ f $(but Arzela-Ascoli theorem tell us that $ f_{k_n}->f $ uniformly . So:$ f_n \to f $ uniformly). and you have that $ \int_0^1 f_n(x) \, dx \to \int_0^1 f(x) \, dx $.