where the action is taken by integration ?
for example, $f=e^{-x}$ is in $L^p$ thus in $L^1$ thus in $L^{1}_{loc}$ and defines a tempered distribution. But what about $g=e^x$ ? Is it true that $g \in L^{1}_{loc}$ since $g$ is continuous and any continuous function is in $L^{1}_{loc}$ ? Then does $g$ define a (tempered) distribution ?
The function $g(x)= e^x$ does not define a tempered distribution. Take $1\geq\psi\geq 0$ to be a smooth function with support in $(-2;2)$ and $\psi=1$ on $(-1,1)$, then $\phi(x) = e^{-\vert x \vert/2} (1-\psi(x))\in \mathcal{S}$. However, $$ \int_\mathbb{R} \phi(x) g(x) dx \geq \int_2^\infty e^{x/2} dx = \infty $$ Thus, $g$ does not define a tempered distribution. However, it does define a distribution (i.e. continuous linear function on the test functions) as it is locally integrable.