Denote uniform norm: $$\|f\|:=|f(t)|_{t\in\mathbb{R}}:=\sup_{t\in\mathbb{R}}|f(t)|$$
Consider a sequence: $$f_n\in\mathcal{C}(\mathbb{R}):\quad\|f-f_n\|\stackrel{n\to\infty}{\to}0$$
Then does it hold: $$\frac{1}{T}\int_0^Tf_n(t)\stackrel{T\to\infty}{\to}M_n\implies\frac{1}{T}\int_0^Tf(t)\stackrel{T\to\infty}{\to}M$$ I suppose this is wrong?
For readability: $M:=\lim_{n\to\infty}M_n$
For all $n$ and $T$, we have $$ \begin{align} \left\lvert \frac{1}{T}\int_{0}^T f - M \right\rvert &\leq \left\lvert \frac{1}{T}\int_{0}^T f - \frac{1}{T}\int_{0}^T f_n \right\rvert + \left\lvert \frac{1}{T}\int_{0}^T f_n - M_n \right\rvert + \left\lvert M_n - M \right\rvert \\ &\leq \frac{1}{T}\int_{0}^T \left\lvert f - f_n \right\rvert + \left\lvert \frac{1}{T}\int_{0}^T f_n - M_n \right\rvert + \left\lvert M_n - M \right\rvert \\ &\leq \lVert f-f_n\rVert_\infty + \left\lvert \frac{1}{T}\int_{0}^T f_n - M_n \right\rvert + \left\lvert M_n - M \right\rvert \\ \end{align} $$
Now, fix $\varepsilon > 0$, and choose $N=N_\varepsilon$ such that for any $n\geq N$ it holds that both $\lvert M_n - M \rvert\leq \frac{\varepsilon}{3}$ and $\lVert f-f_n\rVert_\infty\leq \frac{\varepsilon}{3}$. Then, for any $T>0$ and $n\geq N$,
$$ \begin{align} \left\lvert \frac{1}{T}\int_{0}^T f - M \right\rvert &\leq \frac{\varepsilon}{3} + \left\lvert \frac{1}{T}\int_{0}^T f_n - M_n \right\rvert + \frac{\varepsilon}{3} \\ \end{align} $$ Let $n^\ast\geq N_\varepsilon$ be fixed from now. Using the hypothesis on the convergence of $\frac{1}{T}\int_0^T f_n^\ast$ (which holds for any $n$, so in particular for $n^\ast$): there exists $T_\varepsilon$ such that, for $T\geq T_\varepsilon$, $$ \left\lvert \frac{1}{T}\int_{0}^T f_{n^\ast} - M_{n^\ast} \right\rvert \leq \frac{\varepsilon}{3}. $$ Plugging it back, we have that for any $T\geq T_\varepsilon$, $$ \begin{align} \left\lvert \frac{1}{T}\int_{0}^T f - M \right\rvert &\leq \frac{\varepsilon}{3} + \frac{\varepsilon}{3} + \frac{\varepsilon}{3} = \varepsilon. \end{align} $$ This shows that indeed $\frac{1}{T}\int_{0}^T f \xrightarrow[T\to\infty]{} M$.
Edit: as asked (and answered) in the comments, the assumption that the sequence $(M_n)_n$ converges is not necessary. Specifically, convergence of $(M_n)_n$ towards a limit $M\in\mathbb{R}$ is implied by the other hypotheses: that is, the convergence of $\lVert f- f_n\rVert_\infty$ can be used to show that $(M_n)_n$ is Cauchy.