I would like to show that the discrete mean ergodic theorem implies a continuous version. The discrete mean ergodic theorem I am working with is as follows:
Let $U$ be a unitary operator on a Hilbert space $\mathcal H$ and let $P$ be the orthogonal projection onto the set of $\psi \in \mathcal H$ satisfying $U\psi = \psi$. Then, for any $f \in \mathcal H$, $$\lim_{N \rightarrow \infty} \frac{1}{N} \sum_{n=0}^\infty U^nf = Pf.$$
In the continuous case, we have that $T_t$ is a one-parameter group of measure-preserving transformations and $Uf(w) = f(T_1w)$. We would like to show a mean ergodic theorem for $$\lim_{T \rightarrow \infty} \frac{1}{T}\int_0^T f(T_tw)dt.$$
I have the following approach: Let $(Uf)(w) = f(T_1w)$ and let $\tilde{f}(w) = \int_0^1 f(T_tw)dt$. Note that $$\frac{1}{T} \int_0^T f(T_tw)dt = \frac{1}{T}\sum_{n=0}^{T-1} U^n\int_0^1 f(T_tw)dt = \frac{1}{T} \sum_{n=0}^{T-1} U^n\tilde{f}.$$ By the discrete mean ergodic theorem, this converges to $P\tilde{f}$, the projection of $\tilde{f}$ onto the space $\{g : Ug = g\}$. So, it remains to show that $P\tilde{f} = Pf$.
How can we show this final step? I see that we can understand $\tilde{f}$ as averages of $f$, but I'm not sure what I can do with that.
There are some assumptions missing in question (e.g. where f lives?, it is impossible to integrate a vector in an arbitrary Hilbert space), but this proof should work with these assumptions (if not please comment):
Write $f= f_1 + f_2$ where $f_1$ is $T$-invariant (i.e. $T_t f_1 = f_1$ for all $t$), and $f_2$ is orthogonal to the space of $T$-invariant functions
Then $\tilde{f} = \int_0^1 f_1(T_tw) dt + \int_0^1 T_t f_2(T_tw)dt = f_1 + \int_0^1 f_2(T_tw)dt$.
It is left to show that $\tilde{f}_2 :=\int_0^1 f_2(T_tw)dt$ is orthogonal to the invariant functions.
Let $g$ be an invariant function, then $\left<\tilde{f}_2,g\right>=\int \tilde{f}_2(w) g(w) dw = \int \int_0^1 f_2(T_tw)dt g(w) dw $
By Fubini (and you need some assumptions to justify Fubini here, your question lack these assumptions)
$$\left<\tilde{f}_2,g\right> = \int_0^1 \int f_2(T_tw) g(w) dw dt$$
Since $T_t$ is measure-preserving this equals to $$\int_0^1 \int f_2(w) g(T_{-t} w) dw dt$$ and finally, since $g$ is invariant this equals to $\left<f_2 ,g\right>=0$, as required.