[proofplan] We write the contour integral using the parametrization $\gamma$, so the problem becomes an estimate for a one-dimensional integral over $[a,b]$. The hypothesis $|f(z)| \le M$ on the image of $\gamma$ gives a pointwise bound for the integrand $f(\gamma(t))\gamma'(t)$. Integrating this pointwise bound gives exactly $M$ times the length of the path. [/proofplan] [step:Express the contour integral through the parametrization] Since $\gamma: [a,b] \to \Omega$ is piecewise $C^1$ and $f: \Omega \to \mathbb{C}$ is continuous, the function \begin{align*} h: [a,b] &\to \mathbb{C} \\ t &\mapsto f(\gamma(t))\gamma'(t) \end{align*} is piecewise continuous, with $\gamma'(t)$ understood on the finitely many $C^1$ subintervals of the path. Hence $h$ is integrable with respect to $\mathcal{L}^1$ on $[a,b]$, and the contour integral is defined by \begin{align*} \int_\gamma f(z)\,dz = \int_a^b f(\gamma(t))\gamma'(t)\,d\mathcal{L}^1(t). \end{align*} Therefore \begin{align*} \left|\int_\gamma f(z)\,dz\right| = \left|\int_a^b f(\gamma(t))\gamma'(t)\,d\mathcal{L}^1(t)\right|. \end{align*} [/step] [step:Bound the absolute value of the parametrized integrand] For every $t \in [a,b]$ at which $\gamma'(t)$ exists, we have $\gamma(t) \in \gamma([a,b])$. The hypothesis gives $|f(\gamma(t))| \le M$, and multiplicativity of the complex absolute value gives \begin{align*} |f(\gamma(t))\gamma'(t)| = |f(\gamma(t))|\,|\gamma'(t)| \le M|\gamma'(t)|. \end{align*} The exceptional set where $\gamma'$ is not defined is finite, hence has $\mathcal{L}^1$-measure zero, so the inequality holds $\mathcal{L}^1$-a.e. on $[a,b]$. [guided] The hypothesis controls $f$ only on the trace of the path, namely the set $\gamma([a,b]) \subset \Omega$. This is exactly the set of points sampled by the parametrized integral. For each $t \in [a,b]$ where the derivative $\gamma'(t)$ exists, the point $\gamma(t)$ belongs to $\gamma([a,b])$, so \begin{align*} |f(\gamma(t))| \le M. \end{align*} The integrand in the parametrized contour integral is the product $f(\gamma(t))\gamma'(t)$. Since both factors are complex numbers, the complex absolute value is multiplicative, and therefore \begin{align*} |f(\gamma(t))\gamma'(t)| = |f(\gamma(t))|\,|\gamma'(t)| \le M|\gamma'(t)|. \end{align*} Because $\gamma$ is piecewise $C^1$, there are only finitely many points at which $\gamma'$ may fail to exist. A finite subset of $[a,b]$ has $\mathcal{L}^1$-measure zero, so this pointwise estimate holds $\mathcal{L}^1$-a.e. on $[a,b]$. That is the correct level of precision for an integral estimate. [/guided] [/step] [step:Integrate the pointwise bound and identify the length] Using the integral triangle inequality for the complex-valued integrable function $h$, we obtain \begin{align*} \left|\int_a^b f(\gamma(t))\gamma'(t)\,d\mathcal{L}^1(t)\right| \le \int_a^b |f(\gamma(t))\gamma'(t)|\,d\mathcal{L}^1(t). \end{align*} Applying the pointwise bound from the previous step and monotonicity of the [Lebesgue integral](/page/Lebesgue%20Integral) gives \begin{align*} \int_a^b |f(\gamma(t))\gamma'(t)|\,d\mathcal{L}^1(t) \le \int_a^b M|\gamma'(t)|\,d\mathcal{L}^1(t). \end{align*} Since $M$ is constant and $M \ge 0$, \begin{align*} \int_a^b M|\gamma'(t)|\,d\mathcal{L}^1(t) = M\int_a^b |\gamma'(t)|\,d\mathcal{L}^1(t) = M L(\gamma). \end{align*} Combining these estimates yields \begin{align*} \left|\int_\gamma f(z)\,dz\right| \le M L(\gamma), \end{align*} which is the desired inequality. [/step]