Counterexample to Marcinkiewicz I have a version of Marcinkiewicz:
Let $(X,\mu)$ and $(Y,\nu)$ be measure spaces and let $1<p_1 \leq \infty$. Suppose that $T$ is a mapping from $L^1(X,\mu) + L^{p_1}(X,\mu)$ to $\mu$- measurable functions such that
$1) |T(f +g)(y)| \leq |Tf(y)| + |Tg(y)|$
$2) \nu(\{ y:Tf(y) > \lambda \} \leq (A_0/\lambda) \|f\|_1, \, f\in L^1$
$3) \nu(\{ y:Tf(y) > \lambda \} \leq ((A_1/\lambda) \|f\|_{p_1})^{p_1}, \, f\in L^{p_1}$ 
Then for $1 < p <p_1,$
$$ \|Tf\|_p \leq A_p\|f\|p, \,\,\,f\in L^p$$
where $A_p$ depends only on $A_0$ $A_1$, $p$ and $p_1$.
My questions are why cannot $p_1  = 1$? and why is there a strict inequality in $1 < p < p_1$. I cannot figure out any counterexamples.
Can someone help me? 
 A: The version of the Marcinkiewicz Interpolation Theorem that I know states that for a sublinear map $T$ [your condition $1$] from $L^{p_0}(X, \mu) + L^{p_1}(X, \mu)$ [for you, $p_0 = 1$] to the measurable functions on $(Y, \nu)$, which is of weak type $(p_0, q_0)$ [your condition $2$ with $p_0 = 1$, $q_0 = 1$] and weak type $(p_1, q_1)$ [your condition $3$ with $q_1 = q_1$], where $p_0, p_1, q_0, q_1 \in [1, \infty]$ and $p_0 \leq q_0$, $p_1 \leq q_1$, then $T$ is of strong type $(p_t, q_t)$ for all $t \in (0, 1)$ where $$\frac{1}{p_t} = \frac{1-t}{p_0}+\frac{t}{p_1},\qquad \frac{1}{q_t} = \frac{1-t}{q_0} + \frac{t}{q_1}.$$ In your case, $p_0 = q_0 = 1$ and $p_1 = q_1$ so $p_t = q_t = p \in (1, p_1)$ so the conclusion is the same as your conclusion.
The simple fact that you can't have $p = 1$ is that there are operators of weak type $(1, 1)$ but not of strong type $(1, 1)$. One such example is the Hardy-Littlewood maximal operator $f \mapsto Hf$ where $$Hf(x) = \sup_{r>0}\frac{1}{m(B(r, x))}\int_{B(r,x)}|f(y)|dy.$$
The details for all of the above can be found in Folland's Real Analysis. For example, Theorem $3.17$ shows that the Hardy-Littlewood maximal operator is of weak type $(1, 1)$, and exercise $6.43$ shows that in the case of $X = Y = \mathbb{R}$, $\mu = \nu = m$, $\chi_{(0, 1)} \in L^1(\mathbb{R}, m)$, $H\chi_{(0, 1)} \in L^{1,w}(\mathbb{R}, m)$ but $H\chi_{(0,1)} \not\in L^1(\mathbb{R}, m)$.
