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.

  • $\begingroup$ 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. $\endgroup$ – Anguepa Dec 29 '18 at 15:42
  • $\begingroup$ It would be good if we are given a definition of $\|f\|_1.$ Is it $\|f\|_1 = \int_{[a,b]}|f(x)|\,dx?$ $\endgroup$ – Idonknow Dec 29 '18 at 15:53
  • $\begingroup$ @Idonknow Oh sorry. yes, it is $\endgroup$ – 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.

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

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