Continuity of semigroups on $L^2$ and $L^1$: Is this simple proof correct? Let $(X, \mu)$ be a $\sigma$-finite measure space, and $P_t$ a symmetric, Markovian, strongly continuous contraction semigroup on $L^2(X,\mu)$.  (Markovian means that if $f \in L^2$ with $0 \le f \le 1$, then $0 \le P_t f \le 1$ $\mu$-a.e.  In particular $P_t$ is positivity-preserving.)
I would like to show:

Claim: $P_t$ is also a strongly continuous contraction semigroup on $L^1$.

I have found two proofs of this by Silverstein: In his 1974 book Symmetric Markov processes (MathSciNet), there is a somewhat complicated proof (appended below), which appears to be using a generalization of conditional expectation to $\sigma$-finite measure spaces.  Then in Lemma 1.1 of this 1978 paper, there is a simpler proof, but it still uses some specialized uniform integrability results.
My proof, which follows, is much simpler, which naturally makes me suspect it.  So I'd be interested in comments.
EDIT:  Since asking this question, I found that a proof, very similar to my proof below, appears in N. Bouleau and F. Hirsch, Dirichlet Forms and Analysis on Wiener Space (MathSciNet) at Propositions 2.2.1 and  2.4.2
Proof. Showing that $P_t$ extends to a contraction semigroup on $L^1$ is easy.  Note that $L^1 \cap L^2$ is dense in $L^1$.  Suppose $f \in L^2 \cap L^1$; we will show $||P_t f||_1 \le ||f||_1$, so that $P_t$ extends continuously to $L^1$.  By the Markovian property it suffices to consider $f \ge 0$.  Let $A_n$ be a sequence of sets with finite measure such that $A_n \uparrow X$; then for each $n$,
$$\int_X (P_t f) 1_{A_n} = \int_X f P_t 1_{A_n} \le \int_X f = ||f||_1$$
since the Markovian property gives $P_t 1_{A_n} \le 1$.  (Edit: The first equality holds because $P_t$ is assumed to be symmetric.)  By monotone convergence, as $n \to \infty$ the right side goes to $\int_X P_t f = ||P_t f||_1$.  
Now we show the strong continuity.  First let $f \in L^2 \cap L^1$ with $f \ge 0$.  Let $t_n \downarrow 0$.  We have $P_{t_n} f \to f$ in $L^2$; passing to a subsequence we can assume $P_{t_n} f \to f$ a.e.  For each $n$ we have $|P_{t_n} f - f| \le P_{t_n} f + f$, so we mimic the proof of the dominated convergence theorem:
$$\begin{align*}\int_X 2f &= \int \lim \left(f + P_{t_n} f - |P_{t_n} f - f|\right) \\\\
&\le \liminf \left( \int f + \int P_{t_n} f - \int |P_{t_n} f - f| \right) 
&& \text{(Fatou's lemma)}\\\\
&\le \liminf \left( 2 \int f - \int |P_{t_n} f - f| \right) && \text{since }||P_{t_n} f||_1 \le ||f||_1 \\\\
&= 2 \int f - \limsup \int |P_{t_n} f - f|
\end{align*}$$
which, after rearranging, says $\limsup \int |P_{t_n} f - f| = 0$.  We have thus shown $P_t f \to f$ in $L^1$.  Taking positive and negative parts extends this to arbitrary $f \in L^1 \cap L^2$.  Extending to $f \in L^1$ is also easy since $L^2 \cap L^1$ is dense in $L^1$ and each $P_{t_n}$ is a contraction on $L^1$. QED
Intuitively, Fatou's lemma says that $L^1$ convergence can only fail when the limiting function has too little mass (it can never have too much).  But the contraction property says that this does not happen.
Here is Silverstein's 1974 proof, for reference.  The first line is cut off and says "Lemma 1.3.  For $f \in L^1(dx)$".  (Edit: Incidentally, I'm not able to see why the claimed equality $\mathcal{F}_0 \mathcal{F}_t f(X_0) = P_t (1/P_t 1) P_t f (X_0)$ holds.)

 A: This is more a series of rather trivial comments than an answer but for my own convenience I post them like this.
First of all, as I stated in my comment above, the argument looks fine to me. I now went through it line by line several times and I couldn't spot a mistake.


*

*I'm always a bit afraid of reductions to positive and negative parts, so I spelled them out (sorry if that's all trivial for you): If $f \in L^1 \cap L^2$ then we can write $f = f^{+} - f^{-}$ and thus $P_t \geq 0$ gives $P_t f^\pm \geq 0$, so $(P_t f)^{\pm} = P_t (f^{\pm})$ and hence your limit argument in the first paragraph yields $$\|P_{t}f\|_1 = \int P_{t} f^{+} + \int P_t f^- \leq \int f^+  + \int f^- = \|f\|_1$$
as you asserted.

*At the very end of the first paragraph you seem to have mixed up "right side" and "left side". Of course, you could also replace monotone by dominated convergence here.

*As $L^2$-convergence implies convergence in measure (Vitali), I think there is no need to pass to a subsequence of $t_{n} \downarrow 0$ at the beginning of the second paragraph. More precisely, you could argue using the Fatou lemma for convergence in measure, which would make the argument a bit cleaner in my opinion (even if Fatou's lemma in measure is usually obtained from the pointwise a.e. version by exactly the same reduction).

*Again, I have no objection to the decomposition into positive and negative parts and the extension to all of $L^1$ is clear from an $\varepsilon / 3$-argument.
Summing up, I like your argument and I think it's correct.

Unfortunately, I'm not able to explain the equality $\mathcal{F}_0 \mathcal{F}_t f(X_0) = P_t (1/P_t 1) P_t f (X_0)$ in Silverstein.
A: For semigroups on $L^p$ spaces ($1\le p<\infty$), strong continuity follows from weak continuity; see for example chapter IX of Yosida's book on Functional Analysis. Thus, to show that your extension of $P_t$ to $L^1$ is strongly continuous, it suffices to show that $t\mapsto \int g \, P_tf\,d\mu$ is continuous for each $g\in L^\infty$ and each $f\in L^1$.  By density, it is enough to show this for $f$ and $g$ both in $L^1\cap L^\infty\subset L^2$.
The latter continuity  follows immediately because $P_t$ is a contraction semigroup (on $L^1$ as on $L^2$) that is strongly continuous on $L^2$. (The assertion that $P_t$ is an $L^1$ contraction is  the only place that symmetry of $P_t$ is used; it would suffice for $\mu$ to be an invariant measure for $P_t$, or even just sub-invariant.)
