# 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?

• Drawing graphs of $f_n$ for small values of $n$ may be useful. Apr 1 '19 at 16:50
• It's "humps" of unit area, being successively displaced to the right. I do not see, however, how in the limit as the humps are displaced to infinity, the function is zero.
– EEEB
Apr 1 '19 at 16:52

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 $$1=0$$.

• First fixing an $x\in \mathbb{R}$ it means that $f_n$'s are uniformly continuous. Can I safely assume uniform continuity in this case?
– EEEB
Apr 1 '19 at 17:02
• @EEEB $f_n$ cannot be uniformly continuous since it is not even continuous. If you fix an $x\in\mathbb R$ and look at $(f_n(x))_{n=1}^\infty$, note that the preceding is a sequence of $0$s and $1$s that converges to $0$.
– wjm
Apr 1 '19 at 17:06
• For each $x\in \mathbb{R}$ I'm going to have a sequence $(0,0,1,0,0,\dots)$ etc (thanks for the intuition) with the position of the 1 depending on the $x$ I choose. I can see how in this case the limit is zero. My only point of objection is if I pick an $x$ extremely large ($x\to \infty)$, then the 1 can be displaced arbitrarily far away...
– EEEB
Apr 1 '19 at 17:15
• @EEEB That is correct; you see that to every $x$, there corresponds a sequence of $0$s and $1$s that converges to $0$. Extremely large $x$s have corresponding sequences whose finitely many $1$s are extremely far away. Nevertheless, said sequences still converge to $0$. We say therefore that the sequence of functions $(f_n)_{n=1}^\infty$ converges pointwise to the zero function.
– wjm
Apr 1 '19 at 17:20
• @EEEB I edited my post with more details.
– wjm
Apr 1 '19 at 19:17