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

Let $$a,b\in\mathbb{R}$$ such that $$a and $$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.

• That's actually close to the definition of integral (in the Lebesgue sense) of a non-negative measurable function, where you consider simple functions in place of step functions, see here. – Anguepa Dec 29 '18 at 15:42
• It would be good if we are given a definition of $\|f\|_1.$ Is it $\|f\|_1 = \int_{[a,b]}|f(x)|\,dx?$ – Idonknow Dec 29 '18 at 15:53
• @Idonknow Oh sorry. yes, it is – user472520 Dec 29 '18 at 15:57

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.

• I see, thank you. – user472520 Dec 29 '18 at 16:17