Question Let $1\leq p <\infty$ and let $(X, \mathcal{E}, \mu)$ be a measure space with $\mu(X)=1$. Let $g_n\in \mathcal{L}^p$ for $n \in \mathbb{N}$ and let $g \in \mathcal{L}^p$. Fix $\lambda >0$ and consider the statements (a) and (b) below.
(a)$g_n\rightarrow g$ with respect to $\lVert \cdot \rVert_p$.
(b)$\mu \left( \left\{x \in X \bigg\vert \vert g_n(x)-g(x)\vert\geq \lambda \right\} \right)\rightarrow 0$.
Show that (a) implies (b). Does (b) imply (a)?
Attempt: Let $A = \left\{x \in X \bigg\vert \vert g_n(x)-g(x)\vert\geq \lambda \right\} $ $$\lVert g_n-g \rVert_p = \left( \int_{X} \vert g_n- g \vert^{p}d\mu \right) ^{1/p}\geq \left( \int_{A} \vert g_n- g \vert^{p}d\mu \right) ^{1/p}\geq \lambda \mu(A)^{1/p} $$ so that $$ \mu(A)\leq \left( \frac{\lVert g_n -g\rVert_p}{\lambda}\right)^p$$ and right hand side tends to $0$ as $g_n \rightarrow g$ in the $\mathcal{L}^p$ norm. Hence (a) implies (b).
My Question Does (b) imply(a)? How can I show it? Thank you in advance.