Reference Request - Recovering a function from its definite integrals (inverse problem) I'm having a difficult time finding any theory on an inverse problem I've come up against.  Let's say I have an unknown  function $f:[0,1] \rightarrow \mathbb{R}$, and I know $\int_{a}^{b} f$ for some collection $A$ of pairs $(a,b)\in[0,1]^2$.  I'm looking for pointers to any material that discusses condtions on $f$ and $A$ that are sufficient recover (all of? some of?) the values of $f$.  Google searches just keep turning up elementary-calculus-help-type pages.  I'm a beginning graduate student, if it matters.  Thanks.
Edit: I'm actually looking for something broader than I asked for.  I know that recovering the values of $f$ is a lot to ask and is very unlikely unless $A=[0,1]^2$; I'm also looking for approximations to $f$, anything that can be said about its properties/behaviour, etc. when $A\subsetneqq [0,1]^2$.
 A: Here is how you make an inverse problem of this problem: Choose a space $X$ for the function $f$ you are looking for (e.g. $L^2(0,1)$ to work in Hilbert spaces, but other spaces may be more suitable, depending on your needs).
I assume that you only have finitely many definite integrals (since I assume that this is a practical problem where the definite integrals come from measurements). 
Now let us denote your tuples as $(a_1,b_1),\dots (a_N,b_N)$. You forward operator is
$$\newcommand{\RR}{\mathbb{R}}
K:X\to\RR^N
$$
mapping $f$ to the $N$-vector with components $\int_{a_i}^{b_i}f(x)\,dx$. So you are given a vector $g\in\RR^N$ and want some solution to 
$$
Kf = g.
$$
Now you are in business with the standard theory for linear inverse problems.
You have some of the usual problems coming with an inverse problem: Non-uniqueness (the operator is not injective) and probably instability in some sense (depending on you data and values $(a_i,b_i)$). (As far as I see, non-solvability should not be an issue as $K$ should be surjective for meaningful tupels $(a_i,b_i)$).
To deal with non-uniqueness: You may view this as an advantage as you can choose among all solutions of $Kf=g$. To pick one, you can choose  regularization functional $R:X\to [0,\infty]$ and define a minimum-$R$-solution as solution of
$$
\min\{R(f)\mid f\in X,\ Kf=g\}.
$$
From a computational point of view, convex functional $R$ are beneficial and you can choose $R$ to impose some structure on your solution, e.g. $R(f) = \int_0^1 |f'(x)|^2\, dx$ imposes some smoothness (effectively this means that you constrain your solutions to the Sobolev space $H^1$). The most straight-forward choice would be $R(f) = \int_0^1 |f(x)|^2\, dx$ which should produce a linear equality as optimality condition (and you are effectively computing the Moore-Penrose pseudo-inverse). I could say more about regularizing functionals if needed.
If your data vector $g$ is also uncertain, i.e. it may be given by measurement data with an error, you may want to relax your problem and look for solutions of
$$
\min\{R(f)\mid d(Kf,g)\leq\delta\}
$$
for some discrepancy functional $d$ and some value $\delta>0$. Both should be related to the error in your data. Note that this is in some way equivalent to (generalized) Tikhonov regularization which would be solving
$$
\min_f d(Kf,g) + \lambda R(f) 
$$
for some regularization parameter $\lambda>0$. The most simple case of this would be standard Tikhonov regularization in Hilbert spaces:
$$
\min_f \|Kf-g\|_{2}^2 + \lambda\|f\|_{L^2(0,1)}^2
$$
leading to the linear optimality condition
$$
K^*(Kf-g) + \lambda f = 0.
$$
The adjoint operator $K^*:\RR^N\to L^2(0,1)$ is given by
$$
K^*g = \sum_{i=1}^N g_i\chi_{[a_i,b_i]}
$$
(where $\chi_{[a_i,b_i]}$ is the characteristic function of $[a_i,b_i]$).
So the optimality condition is actually
$$
\sum_i \left[\langle f,\chi_{[a_i,b_i]}\rangle - g_i\right]\chi_{[a_i,b_i]} + \lambda f = 0.
$$
This shows that the regularized solution  is also a linear combination of the characteristic functions $\chi_{[a_i,b_i]}$ and thus, we still get a finite dimensional linear problem  for the coefficients.
If you want some smoothness, try $R(f) = \int_0^1 |f'(x)|^2\, dx$. This would give an optimality conditions like
$$
\sum_i \left[\langle f,\chi_{[a_i,b_i]}\rangle - g_i\right]\chi_{[a_i,b_i]} - \lambda f'' = 0
$$
and thus the solution is piecewise quadratic.
A: In general, it appears that hardly anything interesting can be said. E.g., let $A=\{(1/5,3/5),(2/5,4/5)\}$; here, it will be convenient to think of $A$ as a set of (say) open intervals, rather than a set of pairs of endpoints of intervals. 
However, note first that, without loss of generality, for each open interval in $A$, all the intervals (closed, left-open, right-open) with the same endpoints may be assumed to belong to $A$. Next, let us assume that $\int_0^1|f|<\infty$ and that $A$ is a semi-ring (see measures on semi-rings) 
and $[0,1]\in A$. Then the formula $\mu(I):=\int_I f$ for $I\in A$ defines a finite signed countably-additive measure $\mu$ on $A$, which can be uniquely extended to a signed measure $\bar\mu$ on the sigma-algebra $\Sigma$ generated by $A$. 
The measure $\bar\mu$ determines, and is determined by, the conditional expectation $E(f|\Sigma)$ (of $f$ given $\Sigma$), equal the Radon--Nikodym derivative $\dfrac{d\bar\mu}{d\lambda|_\Sigma}$  (with respect to the underlying Lebesgue measure $\lambda$ over $[0,1]$), and this conditional expectation is then precisely all that we can get from the knowledge of the map $A\ni I\mapsto \int_I f$. (One might note that the appearance of the Radon--Nikodym derivative here is in broad agreement with the comment 
"Recovery of function from its integral is called differentiation" by Alexandre Eremenko.)
E.g., if $A$ consists of all intervals with endpoints in the set $\{j/n\colon j=0,\dots,n\}$, then all that we will know is, in essence, the "histogram" of the average values of $f$ over the intervals $[0,1/n],\dots,[1-1/n,1]$, and this "histogram" is the best approximation to $f$ that we can get in this case. 
Extended comment: Dirk suggested an inverse-problem approach. One may note that such an approach will work perfectly well (and, generally, even better) within the above framework of the conditional expectation. Indeed, for a space $X$ of (say) real-valued integrable functions on $[0,1]$ we have the map $X\overset K\to\mathbb R^A$ defined by the formula $Kf:=(\int_I f)_{I\in A}$ for $f\in X$. This map can  be factored as follows:
\begin{equation}
 X\overset{E(\cdot|\Sigma)}\longrightarrow X_\Sigma\overset{K_\Sigma}\longrightarrow\mathbb R^A,
\end{equation}
where $X_\Sigma$ is the set of all integrable $\Sigma$-measurable functions in $X$ and $K_\Sigma$ is the restriction of $K$ to $X_\Sigma$; indeed, by the definition of the conditional expectation/Radon--Nikodym derivative, we have
$K_\Sigma E(f|\Sigma)=Kf$ for all $f\in X$. Thus, instead of $K$, one can deal with its restriction $K_\Sigma$, with the same (or greater) degree of success. In particular, if $A$ is finite, then we have to deal with the finite-dimensional space $X_\Sigma$ instead of the possibly infinite-dimensional space $X$. 
This comment may be viewed as an illustration of what was said previously: that the conditional expectation $E(f|\Sigma)$ is precisely all that we can get from the knowledge of the map $A\ni I\mapsto \int_I f$. 
A: If $A$ is infinite and dense then you can find the function exactly by taking limits. I'll look at the finite case. In the event that the function is a polynomial of degree $n$ then knowing the integral over $n+1$  internally disjoint intervals is enough for recovery.  Without knowing that the function is a polynomial , but given  $n+1$ integrals one could find the unique polynomial of degree $n$ that matches the data. Or the polynomial of degree $n-m$ that best fits.
