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I am working on a problem that states:

Let $f$ be integrable over $\mathbb{R}$ and $\varepsilon > 0$. Show that there is a simple function $\eta$ on $\mathbb{R}$ which has finite support and $\int_{\mathbb{R}} \lvert f - \eta \rvert < \varepsilon$

What does it mean for a simple function $\eta$ to have finite support? I do not need help with the actual problem, just the meaning. Unfortunately, Royden's fourth edition of 'Real Analysis' does not describe this.

Christian Blatter
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user1876508
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4 Answers4

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It should mean

the function vanishes outside a set of finite measure

but not that only finitely many elements in the domain produce a nonzero value for the function.

We are talking about a measure theory book, after all. The main gauge on the size of a measurable set is its measure.

Of course, if the measure being used is the counting measure then it would turn into "only finitely many values of the domain produce nonzero values.

rschwieb
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In the fourth edition Royden's book, it is written (on p. 79):

"It is convenient to say that a function that vanishes outside a set of finite measure has finite support and define its support to be $\{x: \ f(x) \neq 0\}$".

Thus, a measurable function of finite support means a measurable function supprted on a set of finite measure.

Community
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Prakash Muni Bajracharya
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Wikipedia defines support as the subsetset of the domain where the function is not equal to zero (see https://en.wikipedia.org/wiki/Support_(mathematics)).

Therefore, if a real function $f:\ \mathbb{R}\rightarrow\mathbb{R}$ has a finite support, then there's a subset of $\mathbb{R}$ where it is not zero, and everywhere else it is mapped to zero.

E.g. a function $f$ that is equal to $x^{2}+1$ when $|x|<1$ and zero everywhere else, has finite support, since it is non-zero in the set $(-1, 1)$, and zero in $\mathbb{R}\backslash(-1, 1)$. To be precise, the set $(-1, 1)$ is the function's support (supp($f$)).

In your example, you can show that for all functions that vanish at infinity, like $f$, that is $\lim_{x\rightarrow\pm\infty}f=0$, it is true that the intgral $ \int_{a}^{\infty}f\, dx $ decreases as $a$ increases ($\exists\delta>0$ such as this is true $\forall a>\delta$), and the same holds for negative infinity. Thus, try assigning to $\eta$ $ \eta= \left\{ \begin{array}{ll} f, & x\in (-q, q) \\ 0, & x\in \mathbb{R}\backslash(-q, q) \end{array} \right. $

so that the integral of their absolute difference $\int_{\mathbb{R}}|f-\eta|\, dx$ decreases as $q$ increases. Hence, you have a function $\Phi(q)=\int_{\mathbb{R}}|f-\eta|\, dx,\ q>0$, that is decreasing. From this point on, try a theorem (e.g. Bolzano's theorem) to show that $\Phi:\ (0, \infty)\rightarrow(0, \max(\Phi))$, is equal to $\epsilon$ at at least one point.

kon gram
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  • Regarding you example, the interval $(-1,1) \subseteq \mathbb{R}$ is not finite. – Alex Aug 02 '22 at 15:22
  • @Alex it is not finite, but it has finite measure(i.e. 2) and hence it is an example of function with finite support. – Dinoman Mar 09 '23 at 13:38
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Finite support just means that the function's domain has a finite number of values that produce non-zero values in the range. So $f$ being integrable (i.e. finite valued) suggests that you can bound the domain of your simple function to finite support and still get your convergence.

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    That would imply $\int_{\mathbb{R}} \lvert f - \eta \rvert = \int_{\mathbb{R}} \lvert f \rvert$, since the domain where $\eta$ would have a nonnegative value would be a countable set with measure 0. – user1876508 Oct 29 '13 at 03:31
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    In my older edition (2nd), Royden refers to simple functions vanishing outside a set of finite measure. That would be bounded support, as ncmathsadist indicated, but perhaps he's using finite support as a definition for this (bad one, in my view). –  Oct 29 '13 at 10:55