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Suppose $f:\left[0,1\right]\rightarrow R$ is continuous on $\left(0,1\right)$. Prove there is a sequence of step functions $\left\{f_{n}\right\}$ which converge pointwise to $f$ on $\left[0,1\right]$.

John Franks - A (Terse) Introduction to Lebesgue Integration 2009 - 1.5.6 (7), page 13.

But, consider $f=\sin\frac{1}{x}$, with $f=a$ when $x=0$, is continuous on $\left(0,1\right)$. Suppose that, a step function $f_{n}$, is constant on $\left[0,k\right]$, $k<1$. Since, $-1\leq\sin\frac{1}{x}\leq1$ on $\left(0,k\right)$, so $\lim_{n\rightarrow\infty}f_{n}\neq f$

I can't believe this book is wrong. Is my example correct? If it's not, how to prove this problem? Thanks.

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duqu
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    Your $f$ is not continuous for any $a$. The theorem is about the compact interval $[0,1]$. – Nigel Overmars Jan 19 '16 at 12:16
  • Why would your step function $f_n$ be constant on $[0,k]$? Step functions are only required to take a finite number of values and for each of these values it must take that value on a measurable set. But the set ${x\mid f(x)\geq\frac{1}{2}}$ is perfectly measurable for example. – Jorik Jan 19 '16 at 12:19
  • @Jorik: Are you sure that's the definition of "step function" the book referenced in the question uses? I have seem "step function" used to imply various things, from a partition into a finite number of intervals, to a partition into a countable number of measurable sets. – hmakholm left over Monica Jan 19 '16 at 12:32
  • You can only assume that $f_n$ is constant on $(0,k_n]$ (not $[0,k_n]$!), for some $k_n > 0$ that depends on $n$. And then you can make $(f_n)$ converge to $f$ if $(k_n)$ tends to $0$. – TonyK Jan 19 '16 at 13:33
  • @Nigel Overmars: but in this problem, it's only require $f$ is continuous on $\left(0,1\right)$ – duqu Jan 19 '16 at 22:40

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Your example is wrong because the $k$ in $f_n$ being constant on $[0,k]$ is not independent on $n$. So we could for example have $\lim_{n\to\infty}k = 0$.

In fact that's about how the sequence of step function could be constructed, we just make finer and finer subdivisions of the interval. We select representative points in the interval so that $f_n(2^{-n}j) = f(2^{-n}j)$ for $0\le j<2^n$ and make $f_n$ pointwise constant otherwise (with steps in the midpoints between $2^{-n}j$).

Using this construct we would have $f_n(x) = a$ whenever $x<2^{-n-1}$. This doesn't contradict the fact that $\lim_{n\to\infty}f_n(x)\ne a$ for any $x>0$ (since eventually $2^{-n-1}<x$).

The proof uses the fact that for any point $c$ in the interval you can would have $f(x)$ is arbitrarily close to $f(c)$ given that $x$ is near enough to $c$, and therefore the nearest $2^{-n}j$ is near enough if $n$ is large enough and therefore $f_n(c)-f(c) = f_n(2^{-n}j) - f(c) = f(2^{-n}j)-f(c)$ which is that close.

skyking
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  • Thanks for your answer! It solved my confusion. – duqu Jan 19 '16 at 22:23
  • Thanks for your answer! It solved my confusion. In fact, I meant $k$ here is variable (as $k_{n}$). My problem is I could not find direct example before to continue to think about this exercise. And, when if I use example $f=\frac{1}{x}$, I can understand it, but $f=\sin\frac{1}{x}$ puzzled my brain, your example perfectly explain this for me. Thanks! – duqu Jan 19 '16 at 22:30