Inequality involving supremum and integral

Question

Suppose that f is a real positive and integrable function, I'd like to know if the following holds

$\sup_{t \in [0,1]} \int_\mathbb{R} f(x,t) \: dx \leq \int_\mathbb{R} \sup_{t \in [0,1]} f(x,t) \: dx$.

Is there some reference? Thanks

Answer 1

For fixed $t\in[0,1]$, then $f(x,t)\leq\sup_{t\in[0,1]}f(x,t)$, taking integrals both sides we have $\displaystyle\int_{{\bf{R}}}f(x,t)dx\leq\int_{{\bf{R}}}\sup_{t\in[0,1]}f(x,t)dx$, and then taking supremum on the left to all $t\in[0,1]$ we get the result.

One should note that $\sup_{t\in[0,1]}f(x,t)$ is not necessarily measurable, so the assumption about this measurability should be made.

Answer 2

For all $t \in [0,1]$ and all $x \in \mathbb R$ we have $f(x,t) \le \sup_{t \in [0,1]} f(x,t)$,

hence $\int_\mathbb{R} f(x,t) \: dx \leq \int_\mathbb{R} \sup_{t \in [0,1]} f(x,t) \: dx$ for all $t \in [0,1]$.

This gives $\sup_{t \in [0,1]} \int_\mathbb{R} f(x,t) \: dx \leq \int_\mathbb{R} \sup_{t \in [0,1]} f(x,t) \: dx$