For an integrable function $f$, do continuity conditions on its integral affect continuity of $f$?

In this question, whenever I say "integrable" I mean "Riemann-integrable".

The (first) fundamental theorem of calculus states:

if $f\in C[a,b]$, then the function $F(x)=\int_a^x f(t)\, dt$ is continuous in $[a,b]$, differentiable in $(a,b)$ and $F'(x)=f(x)$ in $(a,b)$.

Now, a (sort of?) converse would read:

if $f$ is an integrable function in $[a,b]$ such that the function $F(x)=\int_a^x f(t)\, dt$ is $C^1[a,b]$, then $f$ is continuous.

This is not true, as the answers below put it, it suffices to alter a continuous function on a point, which does not alter the $C^1$-ness of its integral.

Now I ask this: are these all the counterexamples? Does there exist a counterexample which is not of this kind, i.e. a continuous function save for a removable discontinuity?

To clarify: a function $f:[a,b]\to \mathbb{R}$ is said to have a removable discontinuity in a point $c\in [a,b]$ if both one-sided limits (or the one that is defined, in case of $a$ or $b$) exist and are equal.

Solution 1:

Edit: I understood integrability in the Lebesgue sense. The longer first part is the original answer.

I'm rearranging my answer since I'm a bit confused about the actual question, which seems to be:

If $F(t) = \int_{a}^{t} f(x)\,dx$ is a $C^1$-function, is it then true that $f$ is continuous?

The answer to that question is no, since we can always modify $f$ on a null set without changing $F$. The characteristic function of $\mathbb{Q}$, for example, is nowhere continuous but its integral is zero over every interval. These are true discontinuities, not only removable ones.

The point is that $f$ is completely underdetermined in this question. Any modification of $f$ on a null set will yield the same $F$, thus we can't hope for more than $F' = f$ almost everywhere. Amazingly enough, this turns out to be true.

The right question to ask was answered by Lebesgue. The starting point is that for an integrable function $f$ its definite integral $F$ will always be absolutely continuous. The Lebesgue differentiation theorem even tells us that $F$ is almost everywhere differentiable and $F' = f$ almost everywhere.

Lebesgue proved even more:

Assume that $F$ is differentiable almost everywhere. In order for $F$ to be the definite integral of its derivative it is necessary and sufficient that $F$ be absolutely continuous.

This is proved in any decent text on measure theory e.g. Royden or Rudin.

On the other hand, Luzin's theorem implies that a measurable $f$ is continuous on the complement on a set of arbitrarily small measure.

One further point you should be aware of: If $F$ is continuous and differentiable almost everywhere and $F' = 0$ wherever it is defined then $F$ can be non-constant, even monotonic. the standard example for this is called the Cantor-Lebesgue function, or more colorfully the devil's staircase.

Finally, I'd like to point out that it is an awfully subtle business to characterize the functions which are derivatives of everywhere differentiable functions.

Added:

Robert Israel left the following comment to this answer (thank you very much for that!):

According to Lebesgue's criterion, a function is Riemann integrable on $[a,b]$ if [and only if] it is continuous almost everywhere and bounded. To maintain the continuity almost everywhere in the counterexample, the set on which you modify $f$ can be a closed set of measure $0$, e.g. a Cantor set.

The only thing I'd like to add to that is:

Lebesgue's criterion is actually Riemann's criterion

see Über die Darstellbarkeit einer Funktion durch eine trigonometrische Reihe (1854). More precisely, the relevant passage is Section 5 on pages 226 and 227 of Bernhard Riemann's Gesammelte mathematische Werke und wissenschaftlicher Nachlass.

(Hat tip to Roy Smith and Pete L. Clark)

Solution 2:

The actual conclusion is that $f$ is equal almost everywhere (i.e. except on a set of Lebesgue measure 0) to a continuous function, namely $F'$.

Solution 3:

Counterexample: Define $f$ by $f(x)=0$ if $x \in [0,c)$, $f(c)=1$, $f(x)=0$ if $x \in (c,b]$. Then $f$ is integrable, not continuous, and $F$ is identically zero (hence, in particular, smooth).

EDIT: As an interesting counterexample, let $f$ be the function considered here.

Elaborating: The function $f$ defined in the above link is continuous on $\mathbb{R}-\mathbb{Q}$ (i.e., the irrationals) and discontinuous on $\mathbb{Q}$ (i.e., the rationals). Since $f$ is zero on $\mathbb{R}-\mathbb{Q}$, $F$ is identically zero (in particular, smooth). So this is a counterexample with $f$ being discontinuous on a dense set and, moreover, Riemann integrable.

EDIT (more details). The aforementioned function is known as Thomae's function. For convenience, consider its restriction to $[0,1]$ (still denoted $f$). Since $f$ is bounded on $[0,1]$ and its set of discontinuities (is countable and hence) has measure $0$, $f$ is Riemann integrable. The fact that its integral $F(x)=\int_0^x {f(t)\,dt}$ is identically zero on $[0,1]$ is clear from the picture (given in the Wikipedia link). This also follows from the fact that if a function is (proper) Riemann integrable then it is also Lebesgue integrable and both integrals coincide; indeed the latter can be computed as follows: $$ \int_{[0,x]} {f(t)\,dt} = \int_{[0,x] - \mathbb{Q}} {f(t)\,dt} + \int_{[0,x] \cap \mathbb{Q}} {f(t)\,dt} = \int_{[0,x] - \mathbb{Q}} {0\,dt} + 0 = 0, $$ since $[0,x] \cap \mathbb{Q}$ has measure $0$. This (standard) example of $f$ is particularly appropriate here, because $f$ is discontinuous on a dense subset of its domain.

Solution 4:

The answer is "no".

Take any continuous function $f$, and then alter its values on some null set. Then it can have uncountably many discontinuities, and the integral equality will still hold. (Since it was a null set)

Solution 5:

A nice but different sort of converse appears in Henstock-Kurzweil integration theory:

If $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then the derivative $f=F'$ is necessarily Henstock-Kurzweil integrable on $[a,b]$ and $F(x)=\int_a^x f(t)\,dt$ for all $x$ in $[a,b]$.

This theorem is remarkably easy to prove, also. A nice source for this is a recent book by Yee and Vyborny.