Uniform convergence and Integration question

Question

Let $(f_k)$ be a uniformly convergent sequence of bounded functions on $[0,1]$. Suppose that the sequence $(f_k)$ converges uniformly on $[0,1]$ to a function $f$. Determine if

$\lim_{n \to +\infty}\int_0^{1-\frac{1}{n}} f_n(x)\,dx=\int_0^1 f(x)\,dx.$

So far i have proved that $f$ is continuous on the interval, however am confused

I can prove that $\lim_{n \to +\infty}\int_0^{1-\frac{1}{n}} f_n(x)\,dx=\int_0^{1-\frac{1}{n}} f(x)\,dx.$

Does the fact that the interval lies in $[0,1]$ make a difference as $\frac{1}{n}$ of $0,1$ is either $0$ or undefined, and do i use this with the FTC?

Answer 1

Assuming all the functions in question are integrable, it should follow pretty easily from the triangle inequality.

Hint: First, show that the functions must be uniformly bounded (i.e., there is $M$ so that $|f_n(x)|\le M$ for all $n$ and all $x$). Now show the following: $$\left|\int_0^{1-1/n} f_n(x)dx - \int_0^1 f(x)dx\right| \le \int_0^1 |f_n(x)-f(x)|\,dx + \int_{1-1/n}^1 |f_n(x)|\,dx\,.$$