A subset $E\subseteq \mathbb{R}$ is a Lebesgue measurable set when it satisfies Carathéodory's criterion: for every $T\subseteq \mathbb{R}$,Suppose $m^{\ast }(A)=0$ and let $B\subseteq A$, allowing $B=A$. The monotonicity of Lebesgue outer measure follows directly from covers: every cover of a larger set covers a smaller set. Thus $m^{\ast }(B)=m^{\ast }(T\cap B)=0$, while $m^{\ast }(T\setminus B)\le m^{\ast }(T)$. ConsequentlyThe reverse inequality is countable subadditivity of Lebesgue outer measure applied to $T=(T\cap B)\cup (T\setminus B)$. Equality follows, even if $m^{\ast }(T)=\infty$. Therefore $A$ and every subset $B\subseteq A$ are Lebesgue measurable sets of measure zero. This is the completeness of Lebesgue measure, obtained without presupposing measurability of the outer-null set.

Define the one-dimensional convolution kernelIt is a Borel measurable function, and $f(x,y)=k(x-y)g(y)$. The function $k(x-y)$ is Borel measurable by composition with the continuous difference map. Lifting the Lebesgue measurable function $g$ to the second coordinate preserves measurability for the product sigma-algebra, hence also for Lebesgue measure on ${\mathbb{R}}^2$. Products of measurable functions are measurable, so $f$ is Lebesgue measurable.

Near zero, $|\sin t|\le |t|$; away from zero, $|\sin t|\le 1$. ConsequentlyBoth bounds are integrable on their respective regions by power singularity integrability. In particular,

We use the following forms of the two product-integration theorems. On the product of measure spaces equipped with sigma-finite measures, Tonelli's theorem says that a nonnegative measurable function has equal product and iterated integrals, with infinity allowed. Fubini's theorem says that for an absolutely integrable measurable function, its sections are integrable for almost every fixed coordinate, and both orders of iterated integration equal the finite product integral. The Lebesgue measure on each factor is a sigma-finite measure, because the real line is a countable union of finite-length intervals.

Apply Tonelli's theorem first to $|f|$, before presuming finiteness. By translation invariance of Lebesgue measure,Thus $f$ is Lebesgue integrable on ${\mathbb{R}}^2$. The triangle inequality for an integral now gives the requested estimate:This is the L1 norm identity for a translated kernel product followed by a simple bound on the convolution kernel.

There is also a stronger signed conclusion. The integrable convolution kernel $k$ is an odd function, so ${\int }_{\mathbb{R}}k(t)dt=0$. Having established absolute integrability, we may apply Fubini's theorem to $f$ itself:Unlike part (a), the signed cancellation is justified by a finite integral of the absolute value. The constant eight is useful as an L1 norm bound even though this particular signed total vanishes.

Fix $y\ne 0$. The function $x\mapsto f(x,y)$ is continuous on $[-1,1]$, because its denominator is bounded away from zero, and it is an odd function of $x$. Its integral over this symmetric interval is therefore zero. At $y=0$ the entire section is zero, including the specified value at the origin. Thus the inner $x$ integral exists for every $y$ and is identically zero. The same reasoning with $x$ and $y$ exchanged givesTo test Lebesgue integrability in two dimensions, examine the absolute value near the origin. For $0<\epsilon <1$, the annulus $\epsilon <\sqrt{x^2+y^2}<1$ lies inside the square. In polar coordinates,The change of variables formula on this annulus, where the integrand has no singularity, givesThe lower bound tends to infinity as $\epsilon \downarrow 0$. HenceMoreover reflection in either coordinate axis exchanges the positive part and negative part while preserving Lebesgue measure. Both parts therefore have infinite integral, so a signed double Lebesgue integral would involve the undefined expression $\infty -\infty$.

There is no contradiction with Fubini's theorem: its absolute integrability hypothesis is missing. This is an example of finite iterated integrals without joint integrability, with every individual one-dimensional section integrable despite the two-dimensional singularity.

For every fixed $x>0$, $\log (x+n)/n\to 0$, so $f_n(x)\to 0$. To exchange the limit with the Lebesgue integral, we need a common integrable bound over the whole half-line.

For $n\ge 1$, $x+n\le n(1+x)$, soTogether with $|\sin x|\le \min (x,1)$, this givesFor $0<x\le 1$, $\log (1+x)\le x\le 1$, hence $G(x)\le 2x^{1-q}$, which is integrable because $q<2$. For $x\ge 1$,Both terms are integrable because $q>1$:The second equality follows from integration by parts, or the substitution $x=e^t$. Consequently $G$ is a Lebesgue integrable function, and every $f_n$ is Lebesgue integrable.

The dominated convergence theorem states that almost everywhere convergence of measurable functions under one integrable absolute-value bound implies L1 convergence, and in particular convergence of their integrals. Here the pointwise limit is zero and the dominator is $G$, soIn fact we obtain the stronger conclusion $\Vert f_n{\Vert }_1\to 0$. A quantitative version follows by retaining the factors $1/n$ in the bound:The preceding endpoint estimates show $A,B<\infty$, so the L1 norm is $O(\log n/n)$ as $n\to \infty$, with $O$ denoting Big O notation. This is logarithmic reweighting under dominated convergence.

DefineThis is a Lebesgue integrable function: on $(0,1]$ it is bounded by $x^{-1/2}$, whose integral is two, while on $[1,\infty )$ it is bounded by $e^{-x}$, whose integral is $e^{-1}$. Since $|\sin (xy)|\le x|y|$,Thus $h_y$ is Lebesgue integrable for every $y\ge 0$, including $h_0=0$; the same argument actually works for all real $y$.

The mean value theorem for the sine gives $|\sin u-\sin v|\le |u-v|$. Henceand the triangle inequality for an integral givesThus $F$ is a globally Lipschitz function, in particular continuous on $[0,\infty )$, including the endpoint zero.

For the derivative, fix $y>0$ and take nonzero $t$ sufficiently small that $y+t>0$. The difference quotient has the integrandFor every $x>0$, $D_t(x)\to H(x)\cos (xy)$ as $t\to 0$. The same mean value theorem gives the uniform-in-$t$ bound $|D_t(x)|\le H(x)$.

The dominated convergence theorem states that if measurable functions $u_n$ converge almost everywhere to $u$ and $|u_n|\le G$ for a single Lebesgue integrable function $G$, then $u$ is Lebesgue integrable and $\int |u_n-u|\to 0$, hence $\int u_n\to \int u$. Apply it to the difference quotients along every sequence $t_n\to 0$, using the common dominator $H$. The resulting limit is independent of the sequence, soThe right-hand integral is absolutely convergent because its integrand has absolute value at most $H$. This is dominated differentiation under the integral sign, justified by an explicit integrable bound rather than a formal interchange.