Limit of a sequence of step functions Consider, on the measure space of reals, the sequence $f_1,f_2,\dots$ with $f_n(x)=1$, if $n \leq x \leq n+1$, and zero otherwise. It seems that $\int f_n d\mu=1$ for all $n$. However, the claim is that $\lim_{n\to \infty}f_n=0$. This last limit I want to understand.
Do I prove it via the definition of a limit? Namely,
$$
\forall \epsilon >0 \exists n_o\in \mathbb{N}\forall n\geq n_0 (|f_n|<\epsilon),
$$
and try to find an appropriate $n_0=n_0(\epsilon)$? Is there any general way of understanding what the limit of functions is?
 A: We say that $(f_n)_{n=1}^\infty$ converges to $f$ pointwise if for every $\varepsilon>0$ and every $x\in\mathbb R$, there exists an $N_{\varepsilon,x}>0$ such that for every $n\geq N_{\varepsilon,x}$, it is the case that $|f_n(x)-f(x)|<\varepsilon$.
We say that $(f_n)_{n=1}^\infty$ converges to $f$ uniformly if for every $\varepsilon>0$, there exists an $N_\varepsilon>0$ such that for every $x\in\mathbb R$ and every $n\geq N_\varepsilon$, it is the case that $|f_n(x)-f(x)|<\varepsilon$.
Let $f_n$ be as you prescribed and let $f(x)=0$ for every $x\in\mathbb R$.
To see that $(f_n)_{n=1}^\infty$ converges to $f$ pointwise, let $\varepsilon>0$, let $x\in\mathbb R$, and let $N_{\varepsilon,x}=x+1$. Then observe that for every $n\geq N_{\varepsilon,x}$, it is the case that $|f_n(x)-f(x)|=f_n(x)=0<\varepsilon$.
To see that $f_n$ does not converge to $f$ uniformly, let $\varepsilon=1/2$, let $N_{\varepsilon}>0$, and let $x_n\in[n,n+1]$. Then observe that for every $n\geq N_\varepsilon$, it is the case that $|f_n(x_n)-f(x_n)|=1\geq\varepsilon$.
In order to equate$$\lim_{n\to\infty}\int f_n(x)\text{d}x=\int\lim_{n\to\infty}f_n(x)\text{d}x=\int f(x)\text{d}x,$$it is required that $f_n$ converge to $f$ uniformly. So the OP does not suggest that $1=0$.
