I am only concerned with the integral of a measurable function $f : M \to \mathbb{R}^{\ast}_{+}$, where $\mathbb{R}^{\ast}_{+}$ contains all non-negative real numbers and positive infinity. As I understand it, this is defined as $$\sup \left\{ \int s : \text{$s$ is simple and $s\le f$}\right\}.$$

My question is whether the integral is also equal to $$ \inf \left\{ \int s : \text{$s$ is simple and $s\ge f$}\right\}.$$

  • $\begingroup$ simple functions do not take value $+\infty$ so if $f(x_0)=+\infty$ for some $x_0$ then no $s$ exists with $s\geq f$ so that the $\inf$ you mention is over the empty set, hence takes value $+\infty$. This while the integral over $f$ has not necessarily that value. $\endgroup$ – drhab Sep 18 '17 at 18:06


Let $f:(0,1]\to[0,+\infty]$ be prescribed by $x\mapsto x^{-\frac12}$

Then $\int fd\lambda=\left[2x^{\frac12}\right]^1_0=2$ and this is the outcome of the supremum in your question.

However if $s$ is a simple function then its image is finite so that we can find a value $y>0$ such that $s(x)<y$ for all $x\in(0,1]$. But we can also find a value $x_0$ with $f(x_0)=y$.

Conclusion: no simple function $s$ exists that satisfies $f\leq s$.

Then the infimum in your question is the infimum over the empty set hence takes value $+\infty$.

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  • $\begingroup$ And I think this is still the case if we tweak the definition of simple function to allow it to take the value +infinity? $\endgroup$ – RhubarbAndC Sep 18 '17 at 18:23
  • $\begingroup$ Even if you allow $s$ to have the value $+\infty$ it goes wrong. If it has that value on a set with measure $>0$ then $\int s=+\infty$ so the infimum will not be smaller. If it has that value on a set with measure $0$ then on the complement of that set function $f$ takes unlimited high values so cannot be smaller than $s$ for all elements of that set. $\endgroup$ – drhab Sep 18 '17 at 18:28
  • $\begingroup$ That makes a lot of sense. Thanks for putting that straight $\endgroup$ – RhubarbAndC Sep 18 '17 at 18:37
  • $\begingroup$ You are welcome. $\endgroup$ – drhab Sep 18 '17 at 18:42

If said function is measurable then yes, it is. Some textbooks say that a function $f$ is Lebesgue-integrable if and only if $\sup\{s \text{ simple} : s \leq f\} = \inf\{s \text{ simple} : s \geq f\}$.

It all comes down to showing that these definitions of measurable function $f:X \rightarrow \mathbb{R}$ are all equivalent:

  1. $\forall a \in \mathbb{R}$ the set $\{x \in X : f(x) < a\}$ is measurable.

  2. $\forall a \in \mathbb{R}$ the set $\{x \in X : f(x) \leq a\}$ is measurable.

  3. $\forall a \in \mathbb{R}$ the set $\{x \in X : f(x) > a\}$ is measurable.

  4. $\forall a \in \mathbb{R}$ the set $\{x \in X : f(x) \geq a\}$ is measurable.

which is a piece of cake to prove using the property of closure under complement of a sigma algebra.

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  • $\begingroup$ What do you mean by "it all comes down to..."? How do you get from any of those 4 statements to sup {...} = inf {...}? $\endgroup$ – RhubarbAndC Sep 18 '17 at 17:39
  • $\begingroup$ Well...if you can measure the sets in (1) then you can approximate the function $f$ through simple functions that are $\geq a$ on that set, so $s > f$. You do it for every $a \in \mathbb{R}$ and you get the $\sup$ definition. Likewise in the other cases. $\endgroup$ – barmanthewise Sep 18 '17 at 17:49
  • $\begingroup$ I'll take your word for it :) Thanks. $\endgroup$ – RhubarbAndC Sep 18 '17 at 17:52
  • $\begingroup$ @RhubarbAndC That's risky. Especially on sites like this. $\endgroup$ – drhab Sep 18 '17 at 18:19
  • $\begingroup$ Fair enough. I'm just starting maths at uni and I have never been taught measure theory, so I wasn't necessarily expecting to be able to follow the argument. $\endgroup$ – RhubarbAndC Sep 18 '17 at 18:21

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