Is this ""distribution"" $$ u(t) = 0\; \text{ if }\; t\neq 0 $$ $$ u(t) = \infty \; \text{ if } t=0 $$
is integrable (with respect to the Lebesgue measure ) ?
If u(t) is finite at $t=1$ then there is no problem because u is equal to the zero function almost everywhere so it's integrable. But here $u(0)=\infty$ ... can we use the same argument ? This function is not even in $L^1_{loc}$ so we can't associate a regular distribution ?
We don't need to talk about distributions here. This is a simple function, taking its values in $[0,+\infty]$, which is common in measure theory.
To answer your question, yes you can use the same argument. This function is equal to $0$ $\lambda$-almost everywhere, so it's integrable.