Given that $f(x)$ is continuous on the Real line and that $$\int |f(x)|d\mu < \infty$$ is it a necessity that $\lim_{t \to \infty}f(t)=0$?
I tried very much to get a counterexample but cannot come up with one. I suspect that this is parallel to the result for convergent series of positive numbers.
No, it's not. To find a counter-example, first consider the function $$ h(x) = \begin{cases} 1 - |x| &\text{if $x \in [-1,1]$} \\ 0 &\text{otherwise.} \end{cases} $$ You obviously have $\int_{\mathbb{R}} h = 1$. Now we simply sum up countably many shifted and scaled copies of this, i.e. set $$ f(x) = \sum_{k=0}^\infty h\left(2^k(x - 2k)\right) \text{.} $$ Observe that for $k \in \mathbb{N}$, all except maybe one of these terms is zero on $[2k - 1, 2k + 1]$, i.e. we have $$ f(2k + d) = h(2^kd) \quad\text{if } d \in [-1,1] \text{.} $$ In particular, $$ f(2k) = h(0) = 1 \text{ and } f(2k + 1) = h(2^k) = 0 \text{,} $$ and therefore the limit $\lim_{x \to \infty} f(x)$ does not exist.
But since the area of each "tooth" is half the area of the previous "tooth", $f$ is integrable over the whole real line. We have $\int_{\mathbb{R}} h\left(2^k(x - 2k)\right) \,dx = 2^{-k} \int h = 2^{-k}$, and thus $$ \int_{\mathbb{R}} f = \sum_{k=0}^\infty 2^{-k} = 2 \text{.} $$
I suggest that you plot this function $f$ - if you look at the graph, this will all become very obvious! Also note that we could even have made $f$ be $C^\infty$, instead of merely continuous. We'd just have to replace $h$ by some smooth bump function.
And we could have even made $f$ unbounded! We can introduce an additional $k$-dependent factor that scales the $h$-terms of $f$ vertically, and $f$ will still be integrable so long as the vertical scaling factor doesn't increase too quickly. Try, for example, $$ \hat f(x) = \sum_{k=0}^\infty 2^k h\left(2^{2k}(x - 2k)\right) \text{.} $$
