Recall that a lower semi-continuous function $f:\mathbb{R}^{d}\longrightarrow\mathbb{R}$ is super-harmonic if for any $x$ and $r>0$, we have $$f(x)\geq \dfrac{1}{|B(0,r)|}\int_{B(x,r)}f(y)dy,$$ where $B(x,r):=\{y:|x-y|\leq r\}$ is the ball centered at $x$ with radius of $r$ and $|B(x,r)|$ denotes its volume.
I want to show that if $f(\cdot)$ is a super-harmonic function, then so is $f(\cdot)+c$ for any real number $c$.
It is clear that if $f(\cdot)$ is lower semi-continuous at $x_{0}$, then $$\limsup_{x\rightarrow x_{0}}f(x)\leq f(x_{0})$$ so $$\limsup_{x\rightarrow x_{0}}f(x)+c\leq f(x_{0})+c.$$
Then if $$f(x)\geq\dfrac{1}{|B(0,r)|}\int_{B(x,r)}f(y)dy,$$ we must have $$f(x)+c\geq \dfrac{1}{|B(0,r)|}\int_{B(x,r)}f(y)dy+c.$$
But I am wondering why $$c= \dfrac{1}{|B(0,r)|}\int_{B(x,r)}c dy,$$ should it be $$c= \dfrac{1}{|B(0,r)|}\int_{B(0,r)}c dy?$$ but then how could I combine these two integrals to conclude $$f(x)+c\geq \dfrac{1}{|B(0,r)|}\int_{B(x,r)}f(y)+c dy?$$
Thank you so much!
Lebesgue measure is translation invariant. The measure of $B(x,r)$ is same as that of $B(0,r)$.