Recall the definition of the lower Lebesgue integral: suppose that $f:\mathbb{R}^{d}\to [0,+\infty]$ and are not necessarily measurable. Then $$\underline{\int\limits_{\mathbb{R}^{d}}}f(x)\mathrm{d} x=\sup\limits_{0\leq h\leq f,\;h\text{ simple}}\mathrm{simple}\int\limits_{\mathbb{R}^{d}}h(x)\mathrm{d}x$$
It is obvious that if $c\in[0,+\infty)$ then we have $$\underline{\int\limits_{\mathbb{R}^{d}}}cf(x)\mathrm{d}x = c\cdot \underline{\int\limits_{\mathbb{R}^{d}}}f(x)\mathrm{d}x$$ If $f$ is measurable the equality holds if $c=+\infty$. However, I would like to have an example, when this is not true for $c=+\infty$ (of course, then $f$ must be non measurable).
The first idea that I had was to construct a function that is strictly positive everywhere (so that $+\infty\cdot f$ is equal to infinity at some set of positive measure), but at any set of positive measure it has a sequence of arbitrarily small values. So I took $q\in\mathbb{Q}\cap [-1,1]$ and for every $q$ considered shifts of Vitali set. Note that any measurable subset of Vitali set is a null set, but that does not help much, since any measurable $E_{i}$, on which simple function does not change its value, can contain some shift of Vitali set and some subset of other shift. Thus, one can't assign only one value to these shifts; In some sense, the desired function should change faster. I then thought of additive function, that is not continuous (and then its graph is a dense subset of $\mathbb{R}^2$), but I suppose it does not work either.
I would not like to get the answer right away, but any hints are welcome