Let $f \in L^1(\mathbb R^d)$ such that $f$ is continuous at $0$.
Then $f(0) < \infty$.
Is this true? My (failed) proof attempt goes like this:
Assume $f(0) = \infty$, since $f$ is continous at $x=0$, then, given any $M > 0$, there exists $\delta >0 $ such that $f(x) >M$ for every $x \in (-\delta,\delta)$. Then: $$\int_{-\delta}^{\delta}f > 2\delta M$$ But this doesn't necessarily contradict that $f \in L^1(\mathbb R^d)$. Any help will be much appreciated.
What do you mean by continuity of a function with possibly infinite values?
If by that you mean that, for example, if $f(0)=+\infty$, then $f(x)$ is continuous at zero if $$\forall\varepsilon>0\ \exists \delta>0: \quad\forall x\in(-\delta,\delta)\ f(x)>\varepsilon$$ (in one-dimensional case), then $f(x)=\frac{1}{\sqrt{|x|}}*I_{[-1,1]}(x)$, where $I_{[-1,1]}(x)$ is the indicator function, is a counter-example.