if $f$ is lebesgue integrable and additive then $f$ is continuous

Question

How can I prove that if $f$ is Lebesgue integrable in every interval and additive then $f$ is continuous.

I've already seen that list but only measurable functions that imply continuity appear, in this case $f$ is Lebesgue integrable — José Leonel, Dec 17 '20 at 06:46
If something is Lebesgue integrable, then before that it is measurable, because a pre-requisite for Lebesgue integrability is measurability. — Sarvesh Ravichandran Iyer, Dec 17 '20 at 07:04
So, I can prove that for $f$ Lebesgue mesurable? And is there not another way to do it without going through measurability and only with integrability? — José Leonel, Dec 17 '20 at 07:17

score 3 · Accepted Answer · answered Dec 17 '20 at 07:56

3

Suppose that $f$ is locally Lebesgue integable and solves the Cauchy functional equation

$$ \forall x, y \in \mathbb{R} \ : \quad f(x+y) = f(x) + f(y). $$

Then for each given $x$, we have $ f(x) = f(x+t) - f(t) $ for all $t$, and so, integrating both sides for $0 \leq t \leq 1$ gives

$$ f(x) = \int_{x}^{x+1} f(t) \, \mathrm{d}t - \int_{0}^{1} f(t) \, \mathrm{d}t. $$

Now by the absolute continuity of antiderivative, the right-hand side is continuous, and therefore $f$ is continuous.

answered Dec 17 '20 at 07:56

Sangchul Lee

In the first integral why not is $f(x+t)$... and why the antiderivative is continuous? – José Leonel Dec 17 '20 at 08:15
@JoséLeonel, We have $$\int_{0}^{1}f(x+t),\mathrm{d}t\stackrel{x+t\mapsto t}=\int_{x}^{x+1}f(t),\mathrm{d}t.$$ For your second question, this is a standard textbook result, so I would like to refer to any of your textbook on measure theory. (I can't really imagine a single textbook on introductory measure theory that does not cover this fact...) In this particular case, however, we can also apply the dominated convergence theorem to $$\int_{x}^{x+1} f(t),\mathrm{d}t=\int_{\mathbb{R}} f(t)\mathbf{1}_{[x,x+1]}(t) ,\mathrm{d}t$$ to establish continuity. – Sangchul Lee Dec 17 '20 at 08:29
I understand better the dominated convergence theorem. Thankyou very much. I'm very grateful – José Leonel Dec 17 '20 at 09:22

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