showing that $\|f\|_1=\sup \{\int_{[a,b]}\tau(x)dx \mid \tau \text{ step function and } \tau\le f\} $

Question

Let $a,b\in\mathbb{R}$ such that $a<b$ and $f\colon [a,b]\to \mathbb{R}$ a non-negative function. Is then $$\|f\|_1=\sup \{\int_{[a,b]}\tau(x)dx \mid \tau \text{ step function and } \tau\le f\} ?$$ I have seen it to be true under additionally assuming that $f$ is Riemann-integrable. Therefore, I initially tried to find a counterexample (with a non Riemann-integrable function). But then I came across here Lebesgue integrable implies Riemann integrable?, so that the equality probably is true (note that simple functions are more general as step functions!). And now I don't know whether it's true or not (I tend to vote for wrong, but I still don't have a counterexample)) .. I appreciate any help.

Answer 1

I'm afraid this is not true for all (Lebesgue) integrable functions.

Consider the indicator function on $[a,b]\setminus \mathbb{Q}$, namely $\mathbb{I}_{[a,b]\setminus \mathbb{Q}} : [a,b]\rightarrow \{0,1\}$ where $\mathbb{I}_{[a,b]\setminus \mathbb{Q}}(x)=1$ for $x\in [a,b]\setminus \mathbb{Q}$ and $0$ otherwise.

This function only differs from the indicator function on $[a,b]$ on countably many points, so $$ \int \mathbb{I}_{[a,b]\setminus \mathbb{Q}} = \int \mathbb{I}_{[a,b]}=b-a. $$ However clearly any step function $s:[a,b]\rightarrow \mathbb{R}$ with $s\leq \mathbb{I}_{[a,b]\setminus \mathbb{Q}}$ maps into $(-\infty, 0]$.

It is true however for continuous functions on $[a,b]$, since these are Riemann-integrable.