Is it always possible to use Chasles to decompose an integral? Well, I'm in "classe préparatoire" and I always learn that if f is a continuous fonction integrable on [a,b] and if c is in [a,b] then with Chasles relation we have : 
$$
\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx
$$
I would like to know if there are some functions which are integrable on [a,b] but not on [a,c] and [c,b] ? 
Maybe in infinit dimension? Are possible also in finite dimension? 
Thank's before for your answer ! :D
Shadock 
 A: There are different definitions of integrable. Since you're not yet in university, the standard definition would be the Riemann integral.
Claim: A bounded Riemann integrable function on $[a,b]$ is Riemann integrable on any subinterval of $[a,b]$.
The basic point is that Riemann integrability has to do with chopping your interval into tiny pieces and seeing if the upper Riemann sums and the lower Riemann sums converge to the same number as you subdivide more and more. So one extra split at $c$ will not make a difference. 
Discussion:
Let $f: [a,b] \rightarrow \mathbb{R}$ be a bounded function.
Given a partition $P = \{x_0, x_1, \ldots, x_n\}$ of $[a,b]$ with $a = x_0 \leq x_1 \leq \cdots \leq x_n = b$, we set $\Delta x_i = x_i - x_{i-1}$ and
$U(P,f) = \sum_{i=1}^n \Delta x_i \sup_{x \in [x_{i-1},x_i]} f(x)\ $ (the upper Riemann sum)
$L(P,f) = \sum_{i=1}^n \Delta x_i \inf_{x \in [x_{i-1},x_i]} f(x)\ $ (the lower Riemann sum)
(Here $\sup$ is the supremum, or least upper bound (think: max), and $\inf$ is the infimum, or greatest lower bound (think: min).)
Then we say a function $f$, bounded on $[a,b]$, is Riemann integrable on $[a,b]$, or $f \in \mathcal{R}([a,b])$, if $\inf_{\mathrm{partitions}\ P} U(P,f) = \sup_{\mathrm{partitions}\ P} L(P,f)$, in which case this value is what we define to be the value of the integral $\int_a^b f\, dx$.
Claim: $f\in \mathcal{R}([a,b]) \Longrightarrow f \in \mathcal{R}([c,d])$ for $a \leq c \leq d \leq b$.
This claim follows from the following theorems. See Rudin, Principals of Mathematical Analysis for proofs of these (they're not long).
We say $P^*$ is a refinement of $P$ if $P^* \supset P$.
Theorem 1: If $P^*$ is a refinement of $P$ then $L(P,f) \leq L(P^*,f)$ and $U(P^*,f) \leq U(P,f)$.
That is, splitting up our interval at more points can only help bring the upper and lower Riemann sums closer together.
Theorem 2: $f \in \mathcal{R}([a,b])$ if and only if for every $\epsilon > 0$ there exists a partition $P$ such that $U(P,f) - L(P,f) < \epsilon$.
That is, the function is Riemann integrable if and only if for any $\epsilon$ we can find a partition whose upper and lower Riemann sums are within $\epsilon$ of each other. Together with Theorem 1, we can use this to show $f$ Riemann integrable on $[a,b]$ implies $f$ Riemann integrable on $[c,d]$.
Proof of Claim: Suppose $f \in \mathcal{R}([a,b])$. We aim to prove $f \in \mathcal{R}([c,d])$. By Theorem 2 above, it suffices to show there exists a partition $P$ of $[c,d]$ such that $U(P,f) - L(P,f) < \epsilon$.
We have, by Theorem 2 and $f \in \mathcal{R}([a,b])$, that there exists a partition $Q$ of $[a,b]$ such that $U(Q,f) - L(Q,f) < \epsilon$.
Then by Theorem 1 if we let $Q^* = Q \cup \{c,d\}$ we have $U(Q^*,f) - L(Q^*,f) \leq U(Q,f) - L(Q,f) < \epsilon$.
Then we let $P = Q^* \cap [c,d]$. By considering the definitions of $U(P,f)$ and $L(P,f)$ we see that $U(P,f) - L(P,f) = \sum_{i=1}^n \Delta x_i (\sup_{x\in[x_{i-1},x_i]} f(x) - \inf_{x\in[x_{i-1},x_i]} f(x))$ is a sum of nonnegative terms, as is $U(Q^*,f) - L(Q^*,f)$, and each term of the former appears in the sum for the latter.
Thus $U(P,f) - L(P,f) \leq U(Q^*,f) - L(Q^*,f) < \epsilon$.
Thus we've shown the condition in Theorem 2 for the interval $[c,d]$, and so we have $f \in \mathcal{R}([c,d])$.
