For example, if $f$ is integrable on $[0,3]$, is it also integrable on $[1,2]$? I tried thinking of a counterexample but couldn't, since I've only learned what implies integrability but not what integrability implies.
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First note that if $a \leq a'$ and $b' \leq b$ and $g: [a,b] \to \mathbb R$ is $0$ outside of $[a',b']$, then $\int_a^b g(x)\, dx = \int_{a'}^{b'} g(x)\,dx$. This is easily shown using the definition of the Darboux (or Riemann, if you prefer that definition) integral using partitions (tagged partitions for the Riemann definition).
The product of Riemann–Darboux integrable functions are Riemann–Darboux integrable, so in particular you have (in your concrete example) that $\int_1^2 f(x)\, dx = \int_0^3 f(x) \chi_{[1,2]}(x)\, dx$, where $\chi_{[1,2]}$ denotes the characteristic function of the interval $[1,2]$, i.e.
$\displaystyle\qquad \chi_A(x) = \begin{cases} 1 & \text{if } x \in A \\ 0 & \text{if } x \notin A\end{cases}$
kahen
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A function $f:[a,b]\to\mathbb R$ is Riemann-integrable if and only if $f$ is bounded and its set of points of discontinuity has Lebesgue measure zero. It clearly follow that, if $f$ is Riemann-integrable on an interval, then its restriction to a subinterval is also Riemann-integrable.
bof
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If $f$ is integrable on $[0,3]$ then there is a $F:[0,3]\to \Bbb R$ :$F'=f$ on $[0,3]$.The restriction of $F$ on $[1,2]$ let's name it $G$ means that $G'=f$ on $[1,2]$. So if $f$ is integrable on $[0,3]$ is integrable on every connected subset(<=>here is equal to every subinterval) of $[0,3]$
Haha
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4but integrability doesn't imply the existence of an antiderivative right? so isn't there not necessarily F'=f on [0,3]? – Milo Hou Nov 08 '13 at 09:00
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1$F(x)=\int_0^x f(x)dx$ for $x\leq 3$ is an antiderivative – Haha Nov 08 '13 at 09:07
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2That works if $f$ is continuous. In general, integrable functions need not have antiderivatives. – David Mitra Nov 08 '13 at 09:35