My question is: Suppose that $f:X\rightarrow \mathbb{R}$ is lower semicontinuous and M is a real number. Define $f_M:X\rightarrow\mathbb{R}$ by $$ f_M(x) = \min(f(x),M). $$ Prove that $f_M$ is lower semicontinuous.
The definition that I am using is: A function $f$ is lower semi continuous on X if for all $x\in$X and every sequence $x_n\rightarrow x$, we have $$\lim_{n\rightarrow\infty} \inf f(x_n) \geq f(x).$$ Thank you in advance for any help!
A real-valued function is lower semicontinuous if the inverse image of every set of the form $(-\infty,a]$ is closed (this is actually the definition in many real analysis textbooks, in some others an equivalence). Note that if $a>M$ then $f_M^{-1}((-\infty,a])=X$ which is trivially closed. In the other case, if $a<M$ then $f_M^{-1}((-\infty,a])=f^{-1}((-\infty,a])$ which is closed since you are assuming the lower semicontinuity of $f$. Therefore $f_M$ is lower semicontinuous.
With your definition: let $x$ be any element in $X$ and $\{ x_n \}_{n=1}^\infty$ a sequence converging to $x$. Note first that $f(y) \geq f_M(y)$ for every $y\in X$. Now, since $f$ is LSC we have $$ \liminf_{n\to\infty} f(x_n) \geq f(x) \geq f_M(x) \tag{1} $$ and since we also know that $f_M(x)\leq M$ we have $$f_M(x) \leq \min\{M,\liminf_{n\to\infty} f(x_n) \} \leq \liminf_{n\to\infty} ( \min\{M,f(x_n)\} ) = \liminf_{n\to\infty} f_M(x_n) \tag{2} $$ and since this holds for every $x\in X$ we can conclude that $f_M$ is LSC.
Try to justify (2) using the definition of $\liminf$. $$ \liminf_{n\to\infty} f(x_n) = \inf_{n\geq 0} \sup_{k\geq n}f(x_n) $$.
