Let $f : \mathbb{R}\times(0, 1) \to \mathbb{R}$ be a measurable function such that
a) $f(t,\,\cdot\,)\in L^1(0,1)$ for a.e. $t$,
b) there exists $C > 0$ such that for every interval $(a,\, b) \subset (0,\, 1)$ and for every $s \leq t$ there holds $$ \bigg| \int^b_a f(t,x)- f(s,x)\,\, dx\bigg|\leq C(b-a)(t-s).\qquad (1)$$
Prove that, up to redefining $f$ on a subset of $\mathbb{R}×(0, 1)$ of zero measure, there holds $$|f(t, x) − f(s, x)| \leq C(t − s)\qquad\qquad (2)$$ for every $s \leq t$ and every $x \in (0, \, 1)$.
My solution.
Remark: here I use the notation $\mu$ to denote both the one-dimensional and the two-dimensional lebesgue measure.
From a) there exists a subset $B$ of $\mathbb{R}$ s.t. $f(t,\,\cdot\,)\in L^1(0,1)$ for avery $t\in B$ (with $\mu(\mathbb{R}\setminus B)=0$). In $(1)$ I divide by $b-a$ and I take the limit $b\to a$ obtaining, by lebesgue differentiation theorem, $$|f(t, a) − f(s, a)| \leq C(t − s)\quad \text{ for } a \in A\subset (0,\,1)$$ that is $(2)$ (where $\mu((0,1)\setminus A)=0$). I claim that the subset of $\mathbb{R}×(0, 1)$ of zero (bidimensional) measure is $B \times A$. In order to prove that $\mu((\mathbb{R}×(0, 1)) \setminus (B \times A))=0$ I define the partition $X_n=[n,n+1)$ of the real line. Since $$\mu((B \times A) \cap (X_n \times (0,1)))=\mu((B \cap X_n ) \times A))=1\cdot 1=1$$ we get $\mu((X_n \times (0,1))\setminus (B \times A) )=0$. From additivity of the measure we get $$ \mu((\mathbb{R}×(0, 1)) \setminus (B \times A))=\mu\bigg(\bigcup_n\,\big((X_n \times (0,1))\setminus (B \times A)\big)\bigg)=0.$$
Questions and doubts:
It is al correct ?
I am also interested in different kind of proof.
The claim is wrong. Let $a=-b<0$ and $f(t,x):=e^t x$. Then all integrals $\int_a^b f(t,x)=0$, but there is no estimate of the form $|f(t,x)-f(s,x)|\leq C|t-s|$ valid for all $x\in[a,b]$.