Measure of convolution Let $M$ be the Banach space of all complex Borel measures on $R$.The norm in $M$ is $\|\mu\|=|\mu|(R)$,associate to each Borel set $E\subset R$ the set
$$
E_2=\{(x,y):x+y\in E\}\subset R^2
$$
if $\mu$ and $\lambda$, define their convolution $\mu *\lambda$ to the set function
$$
(\mu *\lambda)(E)=(\mu \times\lambda)(E_2)
$$
for every Borel set $E\subset R$, $\mu \times\lambda$ is product of measure
Please prove $\|\mu *\lambda\|\leq\|\mu\|\|\lambda\|$
This is a homework problem in Rudin's book, can you give me some hint? I try to use definition of norm of measure, but it does not work.
 A: This question has a couple of parts to it.  First, I will show that $|\mu \times \lambda| = |\mu| \times |\lambda|$.
To see this, let $d\mu = h_1 \, d|\mu|$ and $d\lambda = h_2 \, d|\lambda|$ be the polar decompositions of $\mu$ and $\lambda$ respectively (so that $|h_1|=|h_2|=1$).  Let $g(x,y) = h_1(x)h_2(y)$.  Then I claim that $d(\mu \times \lambda) = g \, d(|\mu| \times |\lambda|)$.
Given any measurable $A \subset \mathbb{R}^2$, we know that
\begin{align*}
(\mu \times \lambda)(A) &= \int \lambda(A_x) \, d\mu(x)\\
&= \int \lambda(A_x) h_1(x) \, d|\mu|(x)\\
&= \int \left(\int \chi_{A_x}(y) \, d\lambda(y)\right)h_1(x) \, d|\mu|(x)\\
&= \iint \chi_A(x,y) h_1(x)h_2(y) \, d|\lambda|(y) \, d|\mu|(x)\\
&= \iint_A g \, d(|\mu| \times |\lambda|)
\end{align*}
where I have applied Fubini's Theorem in the first and last equalities (which can be done because both $|\mu|$ and $|\lambda|$ are finite, hence $\sigma$-finite measures on $\mathbb{R}$).
Hence, using theorem 6.13 in Rudin (which states that if $\tau$ is a positive measure and $f \in L^1(\tau)$, then the measure defined by $d\sigma = f\, d\tau$ has $|\sigma|(E) = \int_E |f| \, d\tau$) we can conclude that for any measurable $A \subset \mathbb{R}^2$, we have $$|\mu \times \lambda|(A) = \iint_A |g| \, d|\mu|\, d|\lambda| = \iint_A d|\mu|\, d|\lambda| = (|\mu| \times |\lambda|)(A)$$
which is the first part of what we wanted to show.
Next up, I claim that $|\mu * \lambda|(\mathbb{R}) \leq |\mu \times \lambda|(\mathbb{R}^2)$.  We know that $$|\mu * \lambda|(\mathbb{R}) = \sup \sum_{i=1}^\infty |(\mu * \lambda)(E^i)|$$ where the $E^i$ are a partition of $\mathbb{R}$ (i.e. the $E^i$ are pairwise disjoint and $\cup E^i = \mathbb{R}$) and the supremum is taken over all such partitions.  But each partition $\mathbb{R} = \cup E^i$ induces a partition $\mathbb{R}^2 = \cup E_2^i$.  So the set of partitions of $\mathbb{R}^2$ induced by partitions of $\mathbb{R}$ in this way is a subset of the set of partitions of $\mathbb{R}^2$.  Hence, we have $$|\mu * \lambda|(\mathbb{R}) = \sup \sum_{i=1}^\infty |(\mu * \lambda)(E^i)| = \sup \sum_{i=1}^\infty |(\mu \times \lambda)(E_2^i)| \leq \sup \sum_{i=1}^\infty |(\mu \times \lambda)(F^i)|$$ where the final supremum is taken over all partitions $\mathbb{R}^2 = \cup F^i$.  However, this last supremum is exactly the definition of $|\mu \times \lambda|(\mathbb{R}^2)$, so we have the result $|\mu * \lambda|(\mathbb{R}) \leq |\mu \times \lambda|(\mathbb{R}^2)$.
Now apply these two facts (namely, that $|\mu \times \lambda| = |\mu| \times |\lambda|$ and $|\mu * \lambda|(\mathbb{R}) \leq |\mu \times \lambda|(\mathbb{R}^2)$) and obtain that $$\lVert \mu * \lambda \rVert = |\mu * \lambda|(\mathbb{R}) \leq |\mu \times \lambda|(\mathbb{R}^2) = (|\mu| \times |\lambda|)(\mathbb{R}^2) = |\mu|(\mathbb{R}) \cdot |\lambda|(\mathbb{R}) = \lVert \mu \rVert \lVert \lambda \rVert$$
