Distribution theory of finite sum

Question

I was reading about the distribution function of a $\mu-$measurable function $f$, i.e., the function $$\lambda_f(t)=\mu\{x:|f(x)|>t\},$$ and i read that if $f=g+h$ then it satisfies that $$\lambda_f(t)\leq \lambda_g(t/2)+\lambda_h(t/2).$$ I was wondering if it could be proved that for every finite sum of measurable function $f=\sum_{i=1}^n f_i$ it follows that $$\lambda_f(t)\leq \sum_{i=1}^n\lambda_{f_n}(t/(2^n)).$$ Im not able to prove it from the proof for the case $n=2$. Any help will be very appreacited!

Answer 1

Notice that if $\left|\displaystyle\sum_{i=1}^n f_i(x)\right|>t$, at least one of the $f_i$ must verify that $|f_i(x)|>\dfrac{t}{n}$. Otherwise, by the triangle inequality we would have $$\left|\displaystyle\sum_{i=1}^n f_i(x)\right|\leq\displaystyle\sum_{i=1}^n |f_i(x)|\leq\displaystyle\sum_{i=1}^n \dfrac{t}{n}=t$$

Then, $\{x: |f(x)|>t\}\subseteq \displaystyle\bigcup_{i=1}^n \left\{x: |f_i(x)|>\dfrac{t}{n}\right\} $.

Thus, $\lambda_f(t)\leq \displaystyle\sum_{i=1}^n\lambda_{f_i}(t/n)$.

As $2^n\geq n \space\forall n\ge 1$, we have that whenever $|f_i(x)|>t/n$, $|f_i(x)|>t/2^n$, so $\lambda_{f_i}(t/n)\leq \lambda_{f_i}(t/2^n)$ and your desired result follows too.