Fubini's theorem states that an absolutely integrable measurable function on a product of two $\sigma$-finite measure spaces has integrable sections almost everywhere, the corresponding iterated integrals exist, and both orders equal its product integral. In particular, for $f\in L^1({\mathbb{R}}^2)$,To establish integrability of the nonnegative function here, first use Tonelli's theorem: for a nonnegative measurable function the product integral equals either iterated integral with infinity allowed. This avoids assuming the conclusion needed for Fubini's theorem.

If $\alpha \ge -1$, ${\int }_0^{\infty }(1+x+y{)}^{\alpha }dx=\infty$ for every fixed $y$. If $\alpha <-1$, it isThe outer integral is finite exactly when $\alpha +1<-1$, that is, $\alpha <-2$. In that caseFor $-2\le \alpha <-1$ the outer integral diverges, including logarithmic divergence at $\alpha =-2$. Nonnegativity means finiteness of this integral is exactly absolute integrability; after proving it, Fubini's theorem allows either order.

For a nonnegative measurable function, the Lebesgue integral is defined through nonnegative simple functions:The value may be infinite. A real measurable $g$ is Lebesgue integrable when $\int |g|dx<\infty$; then $\int g=\int g^{+}-\int g^{-}$, where $g^{+}=\max (g,0)$ and $g^{-}=\max (-g,0)$.

If $|g|\le h$, every simple function below $|g|$ is also below $h$. Taking the defining suprema proves monotonicity of the Lebesgue integral:Hence $g$ is integrable, and both its positive and negative parts have finite integrals.

The oscillatory function $k$ is measurable because it is continuous off the singleton $\{0\}$ and has a specified value there. The elementary bounds $|\sin u|\le \min (1,|u|)$ give the integrable majorantIndeed, $\int H=2+2{\int }_1^{\infty }{x}^{-3}dx=3$. The comparison just proved gives $\int |k|\le 3$, so $k$ is Lebesgue integrable over the whole real line, despite its oscillations near zero. Its odd symmetry also gives $\int k=0$, now legitimately because absolute integrability has been established.

Euler's totient function isEuler's theorem states that $\gcd (a,n)=1$ implies $\boxed{a^{\phi (n)}\equiv 1(\mathrm{mod}n)}$. For $n>1$, let $r_1,\dots ,r_{\phi (n)}$ be the invertible residue classes modulo $n$. Multiplication by $a$ permutes these classes, since it has an inverse modulo $n$. Multiplying the resulting congruences givesEvery factor is a unit, so the product can be cancelled, proving the theorem. The case $n=1$ is immediate. For distinct prime numbers $p,q$, removing multiples of either prime from $1,\dots ,pq$ gives

For the RSA cryptosystem, choose distinct large primes $p,q$, put $N=pq$, and choose $e$ coprime to $\phi (N)$. Compute $d$ with $ed\equiv 1(\mathrm{mod}\phi (N))$. The public key is $(N,e)$; the private exponent $d$ is retained secretly. Represent a message by $0\le m<N$, encrypt by $c\equiv m^e(\mathrm{mod}N)$, and decrypt by $m^{\prime }\equiv c^d(\mathrm{mod}N)$.

To prove correct decryption even for messages not coprime to $N$, write $ed=1+k(p-1)(q-1)$. Modulo $p$, if $p\mid m$ then $m^{ed}\equiv m\equiv 0$; otherwise Fermat's little theorem gives $m^{ed}\equiv m$. The same reasoning holds modulo $q$. The Chinese remainder theorem therefore gives $m^{ed}\equiv m(\mathrm{mod}N)$, and the chosen representative recovers $m$. Thus decryption works for every message residue, not merely the units.