Cauchy sequences when $p$ is a function Consider $p$ being a positive bounded and measurable function and $\{f_k\}$ a sequence satisfying 
$$\int_{R^d} |f_k(x)|^{p(x)}dx<\infty$$ and $$\lim_{m,j\to \infty}\int_{R^d} |f_j(x)-f_m(x)|^{p(x)}dx=0$$
then there is a $f$ such that 
$$\int_{R^d} |f(x)|^{p(x)}dx<\infty$$ and $$\lim_{m\to \infty}\int_{R^d} |f(x)-f_m(x)|^{p(x)}dx=0.$$
I proved the result for $p$ being a step function but I couldn't extend the result to the general case.
Other way was to use a mimesis of that $L^p$ is complete. 
 A: Let $\varphi\colon \Bbb N\to \Bbb N$ a (strictly) increasing map such that for each integer $n$
$$\int_{\Bbb R^d}|f_{\varphi(n+1)}(x)-f_{\varphi(n)}(x)|^{p(x)}dx\leq \frac 1{4^n}.$$
Let $A_n:=\{x:|f_{\varphi(n+1)}(x)-f_{\varphi(n)}(x)|^{p(x)}>1/2^n\}$. We have $\mu(A_n)\leq \frac 1{2^n}$, hence for almost every $x$, we can find $N(x)$ such that for all $n\geq N(x)$, $|f_{\varphi(n+1)}(x)-f_{\varphi(n)}(x)|^{p(x)}\leq \frac 1{2^n}$. Since $p>0$, the sequence $\{f_{\varphi(n)}\}$ is almost everywhere convergent, say to a function $f$. We know that for $\alpha\in (0,1)$, we have, since $t\mapsto t^{\alpha}$ is sub-additive,
$$|a+b|^{\alpha}\leq |a|^{\alpha}+|b|^{\alpha},$$
and denoting $M:=\sup_{t\in\Bbb R^d}p(t)$, we have when $p\geq 1$, by convexity,
$$|a+b|^{p(x)}\leq 2^{p(x)-1}(|a|^{p(x)}+|b|^{p(x)})\leq 2^{M-1}(|a|^{p(x)}+|b|^{p(x)}).$$
With these inequalities and Fatou's lemma, we can see that $\int_{\Bbb R^d}|f(x)|^{p(x)}$ is finite and $\lim_{n\to +\infty}\int_{\Bbb R^d}|f(x)-f_{\varphi(n)}(x)|^{p(x)}dx=0$. 
The two mentioned inequalities show that the whole sequence converges.
It seems we didn't use any special feature of $\Bbb R^d$, and the result is true if we replace it by any set, and the Lebesgue measure by any positive one.
