lebesgue and riemann integrals are the same for continuous functions on $[a,b]$

Question

I have a proof in front of me which goes as follows, firstly assuming that the function $f \geq 0$ on $[a,b]$. We get a partition $a = x_0 < x_1 <....<x_n = b$ with $x_i - x_{i-1} = (b-a)/2^n$ and define $$ f_n(x) = \inf_{x\in[x_{i-1},x_i]} f(x)$$ for $i = 1,...,n$ and $f_n(a) = m_1$. Then the author states $f_n$ is increasing to $f$. could someone explain why this is?

Answer 1

It is probably the intent of the proof to define

$$f_n(x) = \sum_{i} \inf_{x\in[x_{i-1},x_i]} f(x) 1_{[x_{i-1},x_i]}.$$

Consider a subinterval $I_{n,k} = [a + k(b-a)/2^n, a+ (k+1)(b-a)/2^n].$

Since $I_{n+1,k} \subset I_{n,k}$ and $I_{n+1,k+1} \subset I_{n,k}$ we have

$$\inf_{x \in I_{n+1,k}}f(x)\geqslant \inf_{x \in I_{n,k}}f(x), \\ \inf_{x \in I_{n+1,k+1}}f(x)\geqslant \inf_{x \in I_{n,k}}f(x).$$

This can be used to show that $f_n$ is increasing.

Convergence $f_n \to f$ follows because for any $x$ we have $x \in I_{n,k}$ for some $k$ and $x$ is within $(b-a)/2^n$ of the point at which $f$ is minimum on the interval. Since $f$ is continuous the difference between $f_n(x)$ and $f(x)$ can be made arbitrarly small for sufficiently large $n$.