What does 'supported in $ [a, b] $' mean in this theorem?


Let $X=[a, b] \subset \mathbb{R},$ and let $\mathbb{X}$ be the collection of Borel sets in $X .$ Let $f:[a, b] \rightarrow[0, \infty)$ be continuous and supported in $[a, b] .$ Then $$ \int f d \lambda=\int_{a}^{b} f(x) d x $$ where the LHS denotes the Lebesgue integral and the RHS denotes the Riemann integral.

  • $\begingroup$ The support of a function is defined as $\{ x : f(x) \neq 0 \}$. Supported in $[a,b]$ means that the support is asubset of $[a,b]$. In this case, it means that outside $[a,b]$ the function $f$ is formally extended to be zero. $\endgroup$ – Crostul Dec 6 '19 at 22:58
  • $\begingroup$ @Crostul i see, but why is this needed then? I don't see why we need to extend the function to the whole real only to make it take a zero value on the extension. Are you sure this is the actual meaning? $\endgroup$ – Loli Dec 6 '19 at 23:04
  • $\begingroup$ In the statement you can remove the part about being supported in $[a,b]$ without changing the meaning. It should not be there in my opinion. $\endgroup$ – Captain Lama Dec 6 '19 at 23:14

The support of a function is, more or less, the set on which the function takes non-zero values. The precise definition may vary a little, depending on the author or context (e.g. if we are considering functions on unstructured sets, then the support is precisely the complement of the zero set, while if we are considering functions on topological spaces, then the support is the closure of the complement of the zero set), but in the context of integration, a reasonable inference is that $$ \operatorname{supp}(f) = \overline{ \{x : f(x) \ne 0\} },$$ where $\operatorname{supp}(f)$ denotes the support of $f$.

If $\operatorname{supp}(f) \subseteq E$, then we may reasonably say that the function $f$ is supported in $E$. This implicitly means that $f(x) = 0$ whenever $x \not\in E$. In the context of the identity given in the question, this is important, as (in the context of Lebesgue integration on the real line) the notation $$ \int f \,\mathrm{d}\lambda $$ typically means "the integral of $f$ over the entire real line". By specifying that $f$ is supported in $[a,b]$, the author is, presumably, attempting to avoid the problem of $f$ being non-zero outside of $[a,b]$. Given that $f$ is explicitly only defined on $[a,b]$, this seems redundant, though perhaps the meaning is, as Crostul suggests, that $f$ should be extended to $\mathbb{R}$ by zero.

In any event, the given theorem is awkwardly stated. If it were me, I might rewrite it as

Theorem: Let $f : \mathbb{R}\to\mathbb{R}$ be a continuous function supported in $[a,b]$. Then $$ \int_{\mathbb{R}} f \,\mathrm{d}\lambda = \int_{a}^{b} f(x) \,\mathrm{d}x, $$ where the integral on the left-hand side is the Lebesgue integral, and the integral on the right-hand side is the Riemann integral.

Alternatively, we might suppose that $f : [a,b] \to\mathbb{R}$, and write $$ \int_{[a,b]} f\,\mathrm{d}\lambda \qquad\text{or}\qquad \int_{\mathbb{R}} f\chi_{[a,b]} \,\mathrm{d}\lambda, $$ where $\chi_{E}$ denotes the characteristic function of $E$.

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  • $\begingroup$ thanks, this seems quite plausible. Like you said, it's quite possible the author was somewhat clumsy with notation, since this seems to be a minor theorem (he didn't even provide proof). I'll post an update here next week as i get clarification from him. $\endgroup$ – Loli Dec 6 '19 at 23:51
  • $\begingroup$ @Loli This actually isn't a minor theorem---it is a very important theorem. It asserts that the Lebesgue integral and the Riemann integral agree for all compactly supported continuous functions. From this, it follows fairly directly that whenever the (proper) Riemann integral exists, the Lebesgue integral also exists, and has the same value. This means that the Lebesgue integral is a legitimate extension of the Riemann integral, which is an incredibly important result. On the other hand, it is not a hard result to prove, so many authors omit the proof entirely. $\endgroup$ – Xander Henderson Dec 6 '19 at 23:56
  • $\begingroup$ could you perhaps help me find a proof? a resource in which it's provided? I would try to prove it myself, but i'm little short of time. : ( $\endgroup$ – Loli Dec 7 '19 at 0:00
  • $\begingroup$ @Loli I'm not in the office right now, so I don't have any of my books near me (so I can't give you a reference off the top of my head), and comments are not really a good place to give proofs. However, there should be plenty of proofs on the interwebs. $\endgroup$ – Xander Henderson Dec 7 '19 at 0:06
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    $\begingroup$ This looks like a non-unreasonable proof. $\endgroup$ – Xander Henderson Dec 7 '19 at 0:07

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