Let , $(X,M,\mu)$ be a positive measure space and let $f\in L^1(\mu)$. Let , $X_n=\{x\in X :|f(x)|>n\}$. Then show that , $\displaystyle \lim_{n\to \infty}n\mu(X_n)=0$.
AS, $f\in L^1(\mu)$ so , $f$ is real valued almost everywhere. When $n\to \infty$ then $\displaystyle \mu(X_n)=\mu\left(\{x\in X:|f(x)|>\infty\}\right)=0.$So the given limit is $0$.
Is it correct or something wrong ?
If something wrong then please provide some hints how to prove it.
Your intuition is correct. Use the dominated convergence theorem to prove this. Define a sequence of functions by $$g_n(x):=\chi_{X_n}(x)|f(x)|.$$ Because $f\in L^1(\mu)$ also $g_n\in L^1(\mu)$ for all $n$ and $|g_n|\le |f|$. The sequence $(g_n)$ converges pointwise (for fixed $x$ choose $n>|f(x)|$) to the zero function. So we get $$\lim_{n \rightarrow \infty} \int_{X_n}|f|d\mu=\lim_{n \rightarrow\infty } \int_X g_nd\mu=\int_X \lim_{n \rightarrow \infty} g_nd\mu=0.$$ Then by the definition of $X_n$ $$n\mu(X_n)=n\int_{X_n} d\mu=\int_{X_n} n d\mu\leq \int_{X_n} |f| d\mu \rightarrow 0, n \rightarrow \infty.$$