[proofplan] We reduce the claim to the Markov inequality. For each $\varepsilon > 0$, the monotonicity of $t \mapsto t^p$ gives $\{|f_n - f| \ge \varepsilon\} \subset \{|f_n - f|^p \ge \varepsilon^p\}$. The Markov inequality bounds the superlevel-set measure by $\varepsilon^{-p}\|f_n - f\|_{L^p}^p$, and the $L^p$ convergence hypothesis sends this bound to zero. [/proofplan] [step:Bound the superlevel-set measure via the Markov inequality] Fix $\varepsilon > 0$ and $n \in \mathbb{N}$. Define $E_n := \{x \in X : |f_n(x) - f(x)| \ge \varepsilon\}$. Since $p \ge 1$ and $t \mapsto t^p$ is non-decreasing on $[0, \infty)$, the condition $|f_n(x) - f(x)| \ge \varepsilon$ implies $|f_n(x) - f(x)|^p \ge \varepsilon^p$, so $E_n \subset \{x \in X : |f_n(x) - f(x)|^p \ge \varepsilon^p\}$. The function $|f_n - f|^p : X \to [0, \infty)$ is $\mathcal{A}$-measurable (as the composition of the measurable function $f_n - f$ with the continuous map $t \mapsto |t|^p$) and integrable ($\int_X |f_n - f|^p \, d\mu = \|f_n - f\|_{L^p}^p < \infty$ since $f_n, f \in L^p(X, \mu)$). Applying the [Markov inequality](/theorems/514) to the non-negative measurable function $g := |f_n - f|^p$ with threshold $\lambda := \varepsilon^p > 0$: \begin{align*} \mu(E_n) \le \mu\!\left(\{|f_n - f|^p \ge \varepsilon^p\}\right) \le \frac{1}{\varepsilon^p} \int_X |f_n - f|^p \, d\mu = \frac{\|f_n - f\|_{L^p}^p}{\varepsilon^p}. \end{align*} The first inequality is [monotonicity](/theorems/1081) ($E_n$ is contained in the larger superlevel set), and the second is the [Markov inequality](/theorems/514). [guided] The goal is $\mu(\{|f_n - f| \ge \varepsilon\}) \to 0$ for every fixed $\varepsilon > 0$. Our only quantitative tool is $\|f_n - f\|_{L^p} \to 0$, which controls $\int |f_n - f|^p \, d\mu$. The Markov inequality is the standard bridge between integrals and superlevel-set measures. Why raise to the $p$-th power before applying Markov? Applying Markov directly to $|f_n - f|$ would require control of $\int |f_n - f| \, d\mu$, which only follows from $L^p$ convergence after an additional step (Holder's inequality or Jensen's inequality). Raising to the $p$-th power matches the integrand to the $L^p$ norm directly: $\int |f_n - f|^p \, d\mu = \|f_n - f\|_{L^p}^p$, requiring no additional argument. The Markov inequality requires (i) $g = |f_n - f|^p \ge 0$ is measurable and (ii) $\lambda = \varepsilon^p > 0$. Both hold. The conclusion $\mu(\{g \ge \lambda\}) \le \lambda^{-1}\int g \, d\mu$ yields the desired bound. [/guided] [/step] [step:Send $n \to \infty$ using $L^p$ convergence] By hypothesis, $\|f_n - f\|_{L^p} \to 0$. Since $t \mapsto t^p$ is continuous at $0$ with $0^p = 0$, $\|f_n - f\|_{L^p}^p \to 0$. For the fixed $\varepsilon > 0$: \begin{align*} 0 \le \mu\!\left(\{x \in X : |f_n(x) - f(x)| \ge \varepsilon\}\right) \le \frac{\|f_n - f\|_{L^p}^p}{\varepsilon^p} \to 0 \quad \text{as } n \to \infty. \end{align*} By the squeeze theorem, $\mu(\{|f_n - f| \ge \varepsilon\}) \to 0$. Since $\varepsilon > 0$ was arbitrary, $f_n \xrightarrow{\mu} f$. [/step]