Show that $\{f_n\}_{n=1}^\infty$ is a decreasing sequence but the convergence is not uniform on $[0,1]$

Let $$f_n(x) := \begin{cases} 1 &\text{for x in } \left(0, \frac{1}{n}\right)\\ 0 &\text{x elsewhere in } [0,1] \end{cases}.$$ Show that $\{f_n\}_{n=1}^\infty$ is a decreasing sequence of discontinuous functions that converges to a continuous limit function, but the convergence is not uniform on $[0,1]$.

I don't really know what do for this question. It's not actually homework, I'm just trying to do practice questions before my final. Any help is appreciated.

• It is decreasing because there are only two values, and $f_n^{-1}(1)$ is decrasing sequence of sets.
• It converges pointwise to $0$ on $(0,1)$ because for a fixed $x$, for $n$ big enough you always has $x\notin (0,1/n)$
• $\sup_{x\in (0,1)} |f_n(x)| = 1 \nrightarrow 0$, hence the convergence is not uniform.
• $f^{-1}(\{1\}) = (0, 1/n)$. this means that the function gets the highest value on this interval, which gets smaller and smaller Apr 21 '14 at 23:48
Well, $\sup\limits_{x\in[0,1]}|f(x)-f_n(x)|$ is always equal to...? Hence it doesn't go to zero, and convergence is not uniform.