Define $$f_j(x) := -j1_{[j,j+1]}.$$ Is it true that $$\int f_j \ d \mu = -j \mu([j,j+1]) = -j?$$ with the Lebesgue measure? I know it is true for simple, positive functions, but does it hold for negative too?
Also, is $\lim_{j \rightarrow \infty} f_j = 0$or what? I'm not sure what $-\infty 1_{\varnothing}$ means.
On
Also, note that for measurable function $f \colon \mathbb{R} \to \mathbb{R}$, Lebesgue integral
$$\int f d \mu = \int f^+ d \mu - \int f^- d \mu$$
(usually by definition) when at least one of the two integrals on the RHS is finite, where
$$\begin{aligned} f^+(x)= & \max\{f(x),0\} \\ f^-(x)= & \max\{-f(x),0\}. \end{aligned}$$
Yes, the integral is defined to be linear, so $$ 0=\int 0 \, d\mu = \int (f-f) \, d\mu =\int f \, d\mu + \int (-f) \, d\mu, $$ so $\int (-f) \, d\mu = -\int f \, d\mu.$
For your second question, it depends what you mean. $f_j \to 0$ pointwise, since for each $x$, $f_j(x)=0$ for all $j$ bigger than some $J$ (that depends on $x$). However, the integrals do not converge, and there is no function $f$ so that $$ \int \lvert f-f_j \rvert \, d\mu \to 0, $$ so in this sense (called strong convergence), $f_j$ does not converge.