Let $f_n$ be a sequence of Riemann integrable functions that converge uniformly to $f$ on the compact interval $[a,b]$. Show that $f$ is bounded and Riemann integrable.

I know that bounded functions on compact intervals are exactly those which are Riemann integrable. I am stuck on showing that the limit function is bounded.

• "I know that bounded functions on compact intervals ... are Riemann integrable". This is false. The typical example is the function $$f(x)=\begin{cases} 0 & x \notin \mathbb{Q} \\ 1 & x \in \mathbb{Q} \end{cases}$$ which is bounded on $[0, 1]$ but is not Riemann-integrable there. – Giuseppe Negro Dec 6 '12 at 23:02

You need to use the uniformity condition. Remember that $f_n$ converges uniformly to $f$ if for every $\varepsilon$, there is an $N$ such that if $n > N$, then $|f_n(x) - f(x)| <\varepsilon$ for all $x \in [a,b]$. So the $f_n(x)$ converge to $f(x)$ at the same speed, in some sense. A common way people visualize this is by taking a band of width $2\varepsilon$ around graph of the limit function $f$, i.e. the set $\{(x,y): y \in [f(x)-\varepsilon, f(x)+\varepsilon]\}$. Now if you graph your $f_n$ too, the uniformity condition implies that $f_n$ lies in this band, if $n$ is sufficiently large. But if $f$ isn't bounded, that requires your $f_n$ to be unbounded.
To formalize this, the argument goes something like this. Let $\varepsilon > 0$. Suppose that $f$ is unbounded. Then for every $M$, there is an $x$ such that $|f(x)|>M$. Furthermore, since $f_n$ converges uniformly to $f$, we know that for some $N$, if $n>N$, we have that $|f(x)-f_n(x)|<\varepsilon$. In particular, if $|f(x)|>M$, $|f_n(x)|>M-\varepsilon$. Since such an $x$ exists for all $M$, $f_n$ is unbounded. Contradicts boundedness of $f_n$. You need to do some work to fix the fact that $f_n$ might be unbounded on some discrete subset of $[a,b]$, but that's the idea.
• Wouldn't the only time the $f_n$'s may be unbounded would be for $n < N$. – emka Dec 6 '12 at 22:58
• If the $f_n$ were continuous, yes. I think you're implicitly assuming that, since actually $f$ could be unbounded. Take for example $f(x) = n$ if $x=1/n$ and $f(x)=0$ otherwise. This is an unbounded Riemann integrable function on [0,1]. Take $f_n =f$. That converges uniformly to $f$, but $f$ isn't bounded. – A.Shrestha Dec 7 '12 at 4:40